Combinatorics

Question 1

Combinatorics is

Answer 1

the number of ways of arranging things

Question 2

When considering the number of possible combinations,

Answer 2

its often helpful to think of ordered slots in a line that we can fill

Question 3

Number of ways of picking one card from each of 20 different packs =

Answer 3

52^20

Question 4

Number of outcomes from the throw of five dice and the toss of five coins =

Answer 4

6^5 x 2^5

Question 5

The number of ways of arranging n objects is

Answer 5

n!

Question 6

0! =

Answer 6

1

Question 7

(n+1)! =

Answer 7

(n+1)n!

Question 8

Permutations: it gives us the number of ways of putting n items into k slots =

Answer 8

n!/(n-k)!

Question 9

Choose function -

Answer 9

order does not matter

Question 10

Number of ways of choosing k items from n =

Answer 10

n!/(n-k)!k!

Question 11

Combinations -

Answer 11

order does not matter

Question 12

Permutations -

Answer 12

order does matter

Question 13

Choosing r things from n number of things, repetition allowed, order matters =

Answer 13

n^r

Repetition allowed, order matters

Question 14

Choosing r things from n number of things, repetition not allowed, order matters =

Answer 14

n!/(n-r)!

Question 15

Choosing r things from n number of things, repetition allowed, order doesn’t matter =

Answer 15

(r+n-1)!/r!(n-r)!

Question 16

Choosing r things from n number of things, repetition not allowed, order doesn’t matter =

Answer 16

n!/(n-r)!r!

Question 17

When considering the number of possible values in a sequence,

Answer 17

consider the number of possibilities in each character position, then multiply together.

Question 18

Find the number of possible 4 digit sequences consisting of right and up movements

Answer 18

Theres two possibilities for each of the four positions so there is 2^4 possibiltiies

Question 19

If there are n items and k slots then the maximum number of items in a slot is at least

Answer 19

n/k

Question 20

6 numbers, 4 positions, no repeats:

Answer 20

6!/(6-4)! possible permutations

Question 21

(y^2 + yx + 1)^n –> coefficient of x^3y^5

Answer 21

1 y^2, 3 xy, (n-4)’s ones : (n 1,3,n-4) = (n! / (n-4)! 3!) = 4(n, 4)

Question 22

(a+b)^n

sum of powers of a and b = n therefore 2n

Answer 22

(y^2 + xy)^k total powers =
2*k = 2k

(y^2 + xy)^k : to get x^3 y^5
2k = 5+3 therefore k=4
3 xy terms 1 y^2 term : 4C1 –> (n, 4)

Question 23

(n,r) =

Answer 23

(n, n-r)

Question 24

AABBB

How many variations

Answer 24

5 choices
2 letters
order does not matter
no repeats
5C2