Combinations and Probability

Question 1

Permutations Formula:

Answer 1

If the order of the selection matters, you can use the permutations formula:

Number of permutations (arrangements) of n items = n!

Number of permutations of k items selected from n =

                 nPk = n!/(n-k)!

E.g. How many ways to arrange letters in word ASCENT?

            6! = 6x5x4x3x2x1 = 720

Another way to solve this problem is to use the “draw blanks” approach, which is the easier approach in many high-difficulty permutations question on GMAT:

You draw six blank spots and for each have to decide the number of possibilities in order. Then multiply all together.

\_\_ x \_\_ x \_\_x \_\_ x \_\_x \_\_ 

For our example: There are 6 letters to chose from for first blank, then 5, then 4, then 3, then 2, and then the last letter.

               6x5x4x3x2x1 = 720

It’s possible to solve most permutations problems without the formula but the formula can save you time on some questions.

Question 2

Hybrids of Combinations and Permutations:

Answer 2

Some questions involve elements of both ordered and unordered selection.

E.g. how many ways to arrange letters in word ASSETS?

This is not just a permutations problem because while the order matters where the letters in the words are the order of the S’s themselves doesn’t matter. So, you couldn’t just calculate 6! arrangements as you have to eliminate redundant arrangements caused by the three S’s from the 6! total.

Since there are 3 S’s, there are 3! ways to rearrange those S’s without changing the word. You need to count every group of 3! within the 6! total as only 1 arrangement.

         I.e. 6!/3! = 6x5x4x3x2x1/3x2x1 = 6x5x4= 120

For word REASSESS this would mean:

8!/4!x2! = 8x7x6x5x4x3x2x1/4x3x2x1x2x1 = 8x7x6x5/2 = 840

REMEMBER: When these questions use the phrase “at least,” then solving for the total and then subtracting the undesired combination or outcomes is usually a very efficient and right approach. Of course subtracting means dividing through the number of undesired combinations.

Question 3

Combinations and Permutations:

Answer 3

Combinations and Permutations let you count the number of possible ways to select a subgroup from a larger group.

Combinations: It’s a combinations question if the selection is unordered. That means that the order in the subgroup doesn’t matter. So, you wouldn’t consider a subgroup as different from another one just because the order of the numbers changed.

Permutation (Vertauschung, Zahlenkombi): It’s a permutations question if the selection is ordered.

Question 4

Combinations Formula for Rain Days:

Answer 4

If it rains on exactly 3 days out of 5 you use the combinations formula to calculate the ways in which this could happen:

5C3 = 5!/3!(5-3)! = 5x4x3x2x1/3x2x1x2x1 = 10

Question 5

Probability Formula:

Answer 5

Probability is likelihood that a desired outcome will occur.

Formula:

Probability = Number of desired outcomes/Number of total possible outcomes.

E.g. If there’s a drawer with 12 shirts of which 9 are white the probability of picking a white shirt at random is 9/12.

For probability of one OR another mutually exclusive events to occur, ADD the probabilities of events.

E.g. In drawer with 12 shirts of which 2 are red and 5 are blue what is the probability that the shirt you pick at random is either red or blue?

Probability to pick red shirt: 2/12
Probability to pick blue shirt: 5/12

Probability for EITHER red or blue: 2/12 + 5/12 = 7/12

To find the probability that one AND another of two independent events will occur, multiply the probabilities of the two events.

Hard probability questions on GMAT ask for the probability of a certain outcome after multiple repetitions of the same or different experiments (e.g. a coin tossed several times). There are 5 approaches you can take on these. Look through cards on approach after repetition.

Question 6

Probability of Outcome After Repetitions 1: Multiplication

Answer 6

If you know the probability for each event and need to find the probability of all the events occurring (e.g. probability the first coin lands heads up, second tails up) you can multiply the probabilities of individual events. Caution: Pay attention on how one event effects another one. Order does matter here.

E.g. 3 students chosen at random from class of 6 girls and 4 boys. What’s the probability that all three are girls?

Probability that first student chosen is a girl: 6/10.
Probability that second student chosen is a girl: 5/9
Probability that third student chosen is a girl: 4/8

Probability that all student chosen are girls:
6/10 x 5/9 x 4/8 = 1/6

Question 7

Probability of Outcome After Repetitions 2: Subtraction of Undesired Outcomes

Answer 7

If you can’t readily calculate the probability of the desired outcome but of the undesired outcome, you can subtract that from the total.

REMEMBER: In probability the total of all possible outcomes is always 1.

E.g. Fair coin (fair means that every outcome is equally possible) is flipped 3 times. What is the probability of getting at least one tail?

Desired: 1 tail, 2 tails or 3 tails.

It’s easier here to look at what’s undesired:

Undesired: 3 heads in a row.

I.e. Total - Undesired

  1 - HHH = 1 - 1/2 x 1/2 x 1/2 = 1 - 1/8 = 7/8

(This approach works for many GMAT questions)

Question 8

Probability of Outcome After Repetitions 3: Multiplying by All Permutations of Outcome

Answer 8

If you need to know the probability that an event will occur a certain number of times, but the order of those occurrences doesn’t matter, solve for the PROBABILITY OF DESIRED OUTCOME, then MULTIPLY BY ALL PERMUTATIONS OF THAT OUTCOME.

E.g. A fair coin is flipped 5 times. What is the probability of getting exactly 3 heads.

What you want: 3 heads and 2 tails but order doesn’t matter. Therefore approach 1 of multiplying the different probabilities of each outcome wouldn’t work well here.

Instead:

1. Use approach 1 to figure out probability of one outcome:
   e. g. HHHTT = 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32
2. Use combinations formula to get number of ways you could get that outcome, i.e. you could re-arrange HHHTT.5!/3! x 2! = 5x4x3x2x1/3x2x1x2x1 = 5 x 2 = 10

I.E. there are 10 different ways to get 3 heads and 2 tails. Each one of those outcomes has a probability of 1/32.

3. Multiply to get the probability of getting exactly 3 heads in 5 coin flips:

                     1/32 x 10 = 5/16

Question 9

Probability of Outcome After Repetitions 4: Work Separately With Numerator and Denominator

Answer 9

If like in approach 3 what you want is very specific and order doesn’t matter you can figure out the total number of possibilities and the total number of desired outcomes separately and then put them together in the fraction that makes the probability formula.

Probability: Number of desired Outcomes/Number of total possible outcomes.

E.g. bag has 4 red marbles, 5 blue marbles, 2 green marbles. If 5 marbles are selected one after another without replacement, what is the probability of drawing 2 red marbles, 2 blue marbles and 1 green marble?

1. Calculate possible outcomes by using the combinations formula:11C5 = 11!/5! x 6! = 11x2x3x7 (leave expression as is for now as we will put it into a fraction again later anyway)
2. Calculate desired outcome:a. 2 of the red marbles: 4C2 = 4!/2!x2!
   b. 2 of the blue marbles: 5C2 = 5!/2!x3!
   c. 1 of the 2 green marbles: 2C1 = 2!/1!x1!now multiply to get all desired outcomes:4!/2!x2! x 5!/2!x3! x 2!/1!x1! = 6x10x2
3. Put fraction together:No. desired outcomes/No. of total possible outcomes
   =
   6x10x2/11x2x3x7 = 20/77

Question 10

Probability of Outcome After Repetitions 5: Count Outcomes

Answer 10

If numbers involved are small then just count the number of possible outcomes:

E.g. in 2 coin flips, what’s the possibility of getting exactly one head?

Possible results in those 2 coin flips:

   HH, HT, TH, TT

Of these 4 possibilities 2 have exactly one head. 

                I.e. probability here is 2/4 = 1/2

Question 11

Probability:

Answer 11

Probability measures the likelihood of a certain event occurring. Probability can be expressed as a fraction, a decimal between 0 and 1, or a percentage between 0% and 100%.

E.g. if there are 2 red marbles and 2 blue marbles in a beg the probability of randomly selecting a red marble from the bag is equal to the ratio of red marbles and all marbles in the bag:

      2:4 or 2/4 = 1/2

Question 12

Manhattan - Combinatorics

Answer 12

Main principles of combinatorics derived from these two simple principles that involve making a decision:

1. OR means add. Decision 1 or Decision 2? (add possibilities)
2. AND means multiply. Decision 1 and Decision 2 (multiply possibilities

E.g. how many choices here?

 Steak OR Chicken OR Salmen  AND   Soup OR Salad

    1  +  1  +  1     = 3                      x          1   +   1   = 2
                          3 x 2 = 6 choices.      

When questions ask “How many…” it’s usually a combinatorics question. However many don’t use OR and AND so you have to use critical thinking.

Question 13

Manhattan - Slot Method:

Answer 13

A type of combinatorics question where you have to find # of possibilities for locks for instance. Since this is an AND decision (you need to pick the first digits, the second, the third and the forth, or even more) you have to multiply here.

The strategy is to start with the most restrictive digits (e.g. if first and last digits have to be odd) then move to the other ones, always subtracting one digit from the pool of possibilities as you go through the slots.

Digit 1 x Digit 2 x Digit 3 x Digit 4

E.g. 10 x 9 x 8 x 7 possible combinations

Question 14

Manhattan - Arranging Groups:

Answer 14

Common type of combinatorics question. Asks how many different ways there are to arrange a group.

Rule:
Number of possible ways to arrange n distinct objects, if there are no restrictions is n!

E.g. how many ways to arrange 4 people in 4 chairs in a row?

Use slot method and rule for factorial in combinatorics. Since this is an AND decision (pick person for first chair, and for second chair, and so on) you need to multiply:

Chair 1 x Chair 2 x Chair 3 x Chair 3

4 x 3 x 2 x 1 = 4! = 24 possibilities to arrange the 4 people
in 4 chairs in a row.

Question 15

Manhattan - Common Factorials to memorize

Answer 15

1! = 1
2! = 2x1 = 2
3! = 3x2x1 = 6
4! = 4x3x2x1= 24
5! = 5x4x3x2x1 = 120
6! = 6x5x4x3x2x1 = 720

Question 16

Manhattan - Arrange Groups by Using Anagram Grid:

Answer 16

In more complicated problems where you have to arrange members of a group, you should use an anagram grid. The basic principle is that you can’t just use the factorial formula and get to the solution.

E.g. how many ways to re-arrange letters in word EEL?

According to factorial formula 3! but that’s not true because EEL and EEL are two re-arrangements but they are the same combination. Same for LEE and LEE. They count as one. You have consider this repetition. What you do is you build a fraction with the total factorial in numerator and the factorials of the the repetitions in denominator:

3!/2! = 6/2 = 3 There are three different ways to re-arrange the word.

You should use an Anagram Grid for complicated cases:

E.g. Local club will send 3 representatives to the national conference. If local club has 8 members, how many different groups of representatives could club send?

Draw grid with as many columns as members of the group, so here 8. Use Y and N in row underneath to indicate if being chosen or not. 3 are being chosen, 5 are not.

Anagram Grid:

1 2 3 4 5 6 7 8
Y Y Y N N N N N

So fraction is:

8!/3!5! = 56

Essentially this question is asking how many arrangements can be made from the word YYYNNNNN.

Why you would use the anagram grid and build a fraction instead of just using factorial formula:

Because for example there are 5! ways in which members 4 through 8 could be grouped not to go but essentially, they are just not going so it’s all the same, so these are repetitions that need to be factored in by dividing out. Likewise there are 3! ways in which members 1 through 3 would be arranged to go but it’s all the same, the order of their arrangement doesn’t matter, they are all going.

Question 17

Manhattan: Multiple Groups

Answer 17

In hard combinatorics questions you have to make decisions and arrange groups. You have to make multiple decision and each decision involves arranging different groups.

In these questions make the groups, arrange each group and then multiply. You have to multiply because you have different groups, which means you are making different decisions. So you determine each decision, i.e. arrangement of each group, separately and then multiply the possible arrangements.

The questions will always make it clear that you’re dealing with multiple decisions.

E.g. a fraternity must choose a delegation of 3 senior and 2 junior members. there are 6 seniors and 5 juniors. How many different delegations possible?

Two decisions to be made: group for seniors, group for juniors. Make arrangements separately for each of the two groups. Then multiply the possibilities because you have to choose both groups and AND means multiply.

1. Senior:
   Using an anagram grid you will find:
   6!/3!3! possibilities (because 3! for chosen and 3! for not chosen that need to be taken into account and divided)

6!/3!3! = 20

2. Juniors:
   Using anagram grid you will find:
   5!/2!3! = 10

Total # of possible delegations: 20x10 = 200.

Another example that shows that you have to pay attention to if it’s a AND or OR question:

Yearbook committee has to pick colors for year book. There are 7 colors and committee can select at most 2 colors.

Here the words “at most” make a whole difference. That means they can chose two colors OR just one. It’s an OR question and the possibilities at the end have to be added. So, you calculate for possibilities depending on whether they chose one color or two separately and then add the possibilities:

1. One color: 7 possibilities
   Another way of showing 7!/1!6! because 7! 1! chosen, 6! not chosen.
   7!/1!6! = 7
2. 7!/2!5! (because 2! combinations for chosen, 6! for not chosen)

7!/2!5! = 21

Solution: There are 7 + 21 = 28 possible color schemes.

Question 18

Manhattan - Probability

Answer 18

Probability is chance or likelihood of an event.

Probability is expressed as fraction:

Probability = Number of Desired Outcome/Number of Possible Outcomes

E.g. flip coin, probability of getting head is: 1/2

Probability fraction will always be between 0 and 1 because the number of desired outcomes is smaller than the number of possible outcomes. There are no negative possibilities. Sometimes an event will be impossible in which case your probability will be 0. Certain events have probability of 1.

GMAT may ask you to express probabilities as fractions, decimal or percent. But always start by building the fraction. Later convert into the desired form.

Question 19

Manhattan - Calculate Numerator and Denominator in Probability Fraction Separately

Answer 19

Calculate numerator and denominator in probability questions separately.

Two ways to calculate numerator and denominator:

1. Use an appropriate combinatorics formula if possible
2. Manually count number of outcomes

E.g. Two cubes with faces 1 to 6 are rolled. Probability that faces add up to 8?

Calculation:

1. Find out possible outcomes: each cube has 6 faces so so possible outcomes: 6x6 = 36
2. Find numerator, i.e. desired outcome. There is no combinatorics formula to help so manually count out.
   Possibilities:
   6+2, 6+4, 5+3, 3+5, 4+4 (note we don’t count in 4+4 twice because it’s essentially the same outcome)

so there are 5 combinations for the desired outcome.

Probability is: 5/36.

Question 20

Manhattan - More than one even in Probability - AND vs. OR

Answer 20

Like in combinatorics AND means multiply, OR means add.

E.g. Two coins flipped. Probability that both head?

1/2 x 1/2 = 1/4 because this is coin 1 AND coin 2 are supposed to bring heads.

E.g. Weather report says 40% chance for sun, 25% for rain, 35% for hail. Probability for rain or hail?

This is an OR question, you get either rain or hail.

Probability: 25% + 35% = 60% = 3/5

Question 21

Manhattan - Probability of Sth. Happening + Sth. Not Happening = 1

Answer 21

The probability of something happening plus the probability of it not happening must be 1.

P(A) + P(Not A) = 1

E.g. Each time S plays she has chance of 40% to win. She plays two games. What’s probability that she loses first game and wins second?

This is an AND question because she plays game1 AND game2.

Probability of loosing: 60% because probability of winning is 40% and both have to sum up to 100% (which is 1).

So result: 60% x 40% = 60/100 x 40/100 = 24/100 = 24%.

Question 22

Manhattan - The 1 - x Probability Trick:

Answer 22

In probability questions that ask for probability that “at least” or “at most” something happens, the 1 - x method is usually faster.

You know that the probability that something happens plus the probability that it doesn’t happen adds up to 1. Therefore:

Probability that something happens = 1 - Probability that something doesn’t happen.

H + NH = 1 therefore H = 1 - NH

E.g. bag of equal number of red, green, and yellow marbles. S takes three out and replaces marbles after each pick. Probability that at least one is red?

Question contains language “at least” so the 1 - x method should be faster:

Pick 1: 2/3 probability that not red
Pick 2: 2/3 probability that not red
Pick 3: 2/3 probability that not red

So probability that all three picks are not red:

2/3 x 2/3 x 2/3 = 8/27

Probability that at least one is red:

1 - 8/27 = 19/27.

The more complicated way of calculating this: find probabilities for each scenario (1 is red, 2 are not, 2 are red, one is not; all three are red) and then add the probabilities. Here you have to add instead of multiply because this is an either OR situation. All three things can’t happen, only one combination will happen (either 1 red, 2 red, or 3 red).

chance to pick red marble: 1/3
chance not to pick red marble: 2/3

1 red marble: 1/3 x 2/3 x 2/3 And there are three different combinations for this outcome:

1/3, 2/3, 2/3
2/3, 1/3, 2/3
2/3, 2/3, 1/3

So multiply probability by 3: 1/3 x 2/3 x 2/3 x 3 = 12/27

2 red marbles: 1/3 x 1/3 x 2/3 and again there are 3 different combinations for her to pick out two marbles. So probability:

1/3 x 1/3 x 2/3 x 3 = 6/27

3 red marbles: 1/3 x 1/3 x 1/3 and there is only one combination because it’s all essentially the same so probability:

1/3 x 1/3 x 1/3 = 1/27

So probability that at least 1 red marble is picked

12/27 + 6/27 + 1/27 = 19/27

Question 23

Combinatorics to Find Probability of Possible Draw:

Answer 23

E.g. magician has 5 animals in his hat: 3 doves, 2 rabbits. If he pulls two at random, what is chance that he will draw a matching pair of animals?

You can find the number of possible draws by using anagram grid and combinatorics formula:

5!/2!3! since two animals will be the pair and 3 will not be.

5!/2!3! = 10 so there are 10 possible pairs the magician can draw.

Now you need to find probability for it being a pair. sign each animal with a and b for Rabbits and x, y, and z for Doves:

Ra + Rb
Dx + Dy
Dx + Dz
Dy + Dz

So 4 possibilities for pairs.

Therefore probability is: 4/10 = 40/100 = 40%

Another way of calculating:

Calculate chance for 2 rabbits, then chance for two doves and then add in the end because it’s either OR.

2 Doves:
First Draw: 3/5 chance
Second Draw: 2/4 chance

Combined chance: 3/5 x 2/4 = 6/20 (this is AND because we are combining first AND second draw)

2 Rabbits:
First Draw: 2/5
Second Draw: 1/4

Combined chance: 2/5 x 1/4 = 2/20

Now add chances to find chance that either two doves OR two rabbits are drawn:

6/20 + 2/20 = 8/20 = 2/5 = 40/100 = 40%

Question 24

Disguised Combinatorics:

Answer 24

Many word problems involving words “how many” are combinatorics problems. Also, many combinatorics problems masquerade as probability problems.

Think creatively to find a combinatorics solution to these questions.

Examples:

A’s home and school are in a neighborhood that’s arranged on a grid. Her school is 3 blocks south and 3 blocks east from her home. Her home and school are both on a corner. If she only walks south and east, how many possible routes can she take to school?

Draw grid:

Home     \_\_\_\_\_\_\_
               |    |    |     |
               |\_\_|\_\_|\_\_|
               |    |    |     |
               |\_\_|\_\_|\_\_|
               |    |    |     |
               |\_\_|\_\_|\_\_|  School

You can see that she walks 3 blocks south and 3 blocks east no matter what route she takes. She makes 6 decisions on her way to school, walk south or east. Three of those decisions have to be east, three south, but those decisions can be made in different combinations. So you can arrange an anagram grid and find out that the combinatorics calculation is:

1 2 3 4 5 6
S S S E E E

6!/3! 3! = 20 possible routes.

Question 25

Arrangements With Constraints:

Answer 25

The most complicated combinatorics problems involve a constraint. It’s a usual combinatorics problem, except for the constraint that makes it difficult.

E.g. G, M, P, J, B and C go to the movies and sit next to each other in one row in six adjacent seats. M and J can’t sit next to each other how many possible arrangements are there?

First ignore constraint and calculate all possible arrangements:

6! = 720

Then subtract the number of arrangements in which J and M would sit next to each other. For that use the glue method. Glue M and J together. MJ is now one person and there are effectively 5 different people. So there are 5! different ways they can sit together. But you have consider one more thing. MJ can also sit as JM so you have to multiply 5! by 2:

5! x 2 = 120 x 2 = 240

(The reason why there are so many arrangements for when MJ or JM sit together is because you also have to count in how the other 4 people sit if MJ or JM sit together.)

So there are:

720 - 240 = 480 possible seating arrangements.

Question 26

Domino Effect:

Answer 26

Sometimes the outcome of the first event will affect the probability of subsequent events. Always check if subsequent events are affected by a former event. the first roll of a dice or flip of coin won’t affect the second but the first draw out of a box will, if the item is not replaced before the subsequent draw.

E.g. A box contains 10 blocks, 3 of which are red. If 2 blocks taken out at random and not replaced, what’s probability that they are both red?

First draw: 3/10 probability
Second draw: 2/9 probability

Combined probability: 3/10 x 2/9 = 6/90 = 1/15

Question 27

Domino Effect and Combinatorics:

Answer 27

Some domino-effect problems are difficult because of the sheer number of possibilities involved. But: if all possibilities are equivalent, combinatorics can help.

General Rule:
When you have a symmetric problem (problem with multiple equivalent cases), calculate probability of one case (often by using the domino-effect rule). Then multiply by the number of cases. Use combinatorics to find number of cases if necessary.

E.g. A gumball machine contains 7 blue, 5 green, and 4 red gumballs. If the machine dispenses three gumball at random, what is probability that it’s one gumball of each color?

First consider one possible draw: blue, green, red

Using domino effect you can calculate probability for that:

7/16 x 5/15 x 4/14 = 1/24

Another possible draw: green, blue red

5/16 x 7/15 x 4/14 = 1/24

You will see that all possibilities have the same probability because the order in which the balls come out doesn’t matter and numerator will always have 5x7x4 and denominator always 16x15x14. So it will always be 1/24.

Because the three desired gumballs can come out in any order, there are 3! or 6 cases and they all have the same probability. Therefor you result for overall probability is:

1/24 x 6 = 1/4

This is a truly symmetric situation because if you swapped the order of red and green dispensed by machine, nothing would change about the problem.

Question 28

Remember for Combinatorics and Probability:

Answer 28

In questions that combine combinatorics and probability, you can use two step process:

1. Use combinatorics in the form of:

9!/2!7!

or any fraction like that to find the total number of combinations or possibilities.

REMEMBER for Combinations formula that you are essentially asking for different way so arrange a word, often something in from of:

YYNNNNNNN

or any combination like that. (here Y means yes, N no)

2. Then figure out the probability that the desired outcome happens by counting or using combinations formula again. then build your probability fraction:

Probability = Desired outcome/possible outcome

Or you can right away work with probability tree method where you look at how one event affects a subsequent event and by that put together fractions that you multiply or add in the end.

ALSO remember that often it’s easier to calculate the undesired event and then subtract from 1 in complicated questions that otherwise would require a lot of calculation.

Also pay attention to when order matters and when it doesn’t matter.

Question 29

If You Are Given Number of Combinations And Need to Find Original group.

Answer 29

If GMAT give you the number of combinations possible to form a specific group of a specific number of people among an unknown group then you can find that unknown number.

E.g. if GMAT tells you that there are 56 ways to choose 3 people out of a group x. Then it is mathematically possible to find the size of that group. GMAT doesn’t expect you to know the answer but only to know that you can theoretically find the answer (for DS).