Answer:
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}.
Step-by-step 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}.
