Permutation and Combinatorics

Question 1

If we have 6 people how many permutations are possible? In how many ways can we seat those people?

Answer 1

6 * 5 * 4 ways = 120 permutations

Question 2

If we have 6 people how many combinations of 3 people can we have? The order of people does not matter. So group A,B,C is same as C,B,A?

Answer 2

If we look at all permutations there are 654 ways to choose them = 120 ways.
Now [[A,B,C],[A,C,B],[B,A,C],[B,C,A],[C,A,B],[C,B,A]] are all the same. So it will be treated as one.
So the formula will become
No of permutations / Ways to choose 3 people
= 120/6 = 20 ways.
This is the formula for (6 C 2)

Question 3

Number of ways to choose k people from group of n. Formula for nPk

Answer 3

n!/(n-k)!

We divide by (n-k)! because those permutations will not be possible due to k being smaller than n

Question 4

Number of combinations of k people from group of n. Formula for nCk

Answer 4

of ways to choose k people from group of n / # of ways to choose k people

We divide by # of ways to choose k people because in combinations, A,B,C or C,B,A are counted as one, and there are 6 such permutations.

So nPk / k!
= n!/(n-k)!k!

Question 5

Binomial Co-efficient

Answer 5

n!/(n-k)!k! is called binomial co-efficient. It is the formula for nCk i.e. number of combinations of k people from a group of n people.

Question 6

You have 3 boxes and we want to color it using atmost 4 colors. The only condition is that two boxes with same color should not be placed with each other. How many ways are there to color the boxes?

Answer 6

First we will choose 3 colors out of possible 4. So we have 4C3 combinations. Now with three colors we can make 3 * 2 * 1 permutations. 4C3 * 3!. Now we can also color the board with 2 colors. So we choose 2 colors out of 4, which is 4C2 and there are 2! ways to color the boxes. 4C2 * 2!. Total comes to 4C3 * 3! + 4C2 * 2! = 36

Question 7

What is permutation?

Answer 7

Permutation is ordering of list of objects. For example arranging 4 people in line. It is equivalent to finding permutations of 4 objects. Here in line the ordering is important.

Question 8

Should all objects appear in permutation?

Answer 8

Yes all objects must appear in permutation and two orderings are considered different if some object appears in different place in the orderings.
Like ABC is different than BAC or CBA.

Question 9

If you have n distinct objects, how many different ways of arranging the objects?

Answer 9

n * (n - 1) * (n - 2) * …. * 1 = n!

Question 10

How many ways can we seat 4 people in circular order?

Answer 10

In case of linear arrangement the number of ways is equal to 4!. Now the thing to note here is that ABCD, BCDA, CDAB, DABC are all the same as we can get one from other by rotating. Note that the relative order of characters don’t change. So we can consider them as one.
So the question that we really need to ask is how many groups of family of 4 can we have from 4 people. So it will be 4!/4 = 3!.

This can also be thought of as keeping one person fixed and finding out different ways in which other can be permuted around him. So it will be (4 - 1)! = 3!

Question 11

How many ways can we seat 4 people in circular order irrespective of order, like clockwise and anti clockwise.

Answer 11

Ways to seat people taking order in mind is (4) ! / 4.
Not out of this if we don’t consider order, that is clockwise or anticlockwise, then ABC, ACB and BAC, BCA etc are same. So the number of ways reduces by 2. So we have (n - 1)! / 2 ways. 3! / 2 = 3

Question 12

Practical applications of arranging things in circle without caring about order, clockwise or anticlockwise.

Answer 12

Arranging beads of same color in a necklace.

Question 13

How many ways can we seat 4 people in circular order irrespective of order or no dependency of order, like clockwise and anti clockwise.

Answer 13

Ways to seat people taking order in mind is (4) ! / 4.
Not out of this if we don’t consider order, that is clockwise or anticlockwise, then ABC, ACB and BAC, BCA etc are same. So the number of ways reduces by 2. So we have (n - 1)! / 2 ways. 3! / 2 = 3

Question 14

Practical applications of arranging things in circle without caring about order, clockwise or anticlockwise.

Answer 14

Arranging beads of same color in a necklace.

Question 15

There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?

Answer 15

1800. A really interesting question, multiple solutions possible. https://www.careerbless.com/aptitude/qa/permutations_combinations.php

Question 16

Find out the number of ways in which 6 rings of different types can be worn in 3 fingers?

Answer 16

729.
The first ring can be worn in any of the 3 fingers (3 ways).
Similarly each of the remaining 5 rings also can be worn in 3 ways.
Hence total number of ways
= 3 ^ 6 = 729

Question 17

In how many ways can three boys can be seated on five chairs?

Answer 17

60
The first boy can sit in any of the five chairs (5 ways).
Now there are 4 chairs remaining. The second boy can sit in any of the four chairs (4 ways).
Now there are 3 chairs remaining. The third boy can sit in any of the three chairs (3 ways).
5 * 4 * 3 = 60