Question

(1) Suppose Sanjay has a large pile of identical black marbles and a large pile of identical orange marbles. (a) How many ways could Sanjay set up a line that is 7 marbles long? Write a list (in an organized manner) of 10 outcomes for this problem (b) How many ways could Sanjay set up a line of marbles of length 5 or less? (A line of marbles must contain at least one marble.)

Answer 1

a) N = \frac{n!}{(7-m)!m!}

b) N = N = N(1)+N(2)+N(3)+N(4)+N(5)

N(i) = P(n)(i) = \frac{n!}{i!}, i = {1,2,3,4,5}.

Explanation:

a) Supposing Sanjay has n total marbles, and the order which the marbles are positioned is important(it means that B-B-B-O-B-B-B and O-B-B-B-B-B-B are different outcomes), the number of ways N that Sanjay can set up a 7 marble long line is a permutation of n with repetition(where m is the number of black marbles and 7-m is the number of orange marbles).

Mathematically, this is written as:

N = P(n)(m, 7-m) = \frac{n!}{(7-m)!m!}.

where P means permutations.

10 outcomes for this problem are:

B-B-B-B-B-B-B

B-O-B-B-B-B-B

B-O-O-B-B-B-B

B-O-B-O-B-B-B

B-O-B-B-O-B-B

B-O-B-B-B-O-B

B-O-B-B-O-O-O

B-O-O-B-O-B-B

B-O-O-B-O-O-B

O-O-B-B-O-B-B

...

b) For this question, you have to consider the ways that Sanjay could set up a line with one marble(N(1)), two marbles(N(2)), three marbles(N(3)), four marbles(N(4)) and five marbles(N(5)). The number of ways is N = N(1)+N(2)+N(3)+N(4)+N(5), where

N(i) = P(n)(i) = \frac{n!}{i!}, i = {1,2,3,4,5}.