Bunuel wrote:
There are n identical red balls & m identical green balls. The number of different linear arrangements consisting of “n red ball but not necessarily all the green balls’ is xCy then
(A) x = m + n, y = m
(B) x = m + n + 1, y = ,m
(C) x = m + n + 1, y = m + 1
(D) x = m + n, y = n
(E) x = m + n, y = n + 1
Solution:
If the linear arrangement consists of n red balls and no green balls, then there are n! / n! = nCn ways.
If the linear arrangement consists of n red balls and 1 green ball, then there are (n + 1)! / n! = (n+1)Cn arrangements.
If the linear arrangement consists of n red balls and 2 green balls, then there are (n + 2)! / (n! x 2!) = (n+2)Cn arrangements.
…
If the linear arrangement consists of n red balls and m - 1 green balls, then there are
(n + m - 1)! / (n! x (m - 1)!) = (n+m-1)Cn arrangements.
If the linear arrangement consists of n red balls and m green balls, then there are
(n + m)! / (n! x m!) = (n+m)Cn arrangements.
Now, we need to add up these numbers of arrangements.
nCn + (n+1)Cn + (n+2)Cn + … + (n+m-1)Cn + (n+m)Cn
Of course, we need to simplify this expression. However, if we don’t know how to do that algebraically, we can always do it arithmetically. That is, let’s use easy numbers for n and m. So, let’s say n = 2 and m = 4, we have:
2C2 + 3C2 + 4C2 + 5C2 + 6C2 = 1 + 3 + 6 + 10 + 15 = 35
Now, let’s see which answer choice (or choices) will yield 35 when n = 2 and m = 4:
A) x = 4 + 2 = 6, y = 4 → xCy = 6C4 = 15 → This is not 35.
B) x = 4 + 2 + 1 = 7, y = 4 → xCy = 7C4 = 35 → This IS 35.
C) x = 4 + 2 + 1 = 7, y = 4 + 1 = 5 → xCy = 7C5 = 21 → This is not 35.
D) x = 4 + 2 = 6, y = 2 → xCy = 6C2 = 15 → This is not 35.
E) x = 4 + 2 = 6, y = 2 + 1 = 3 → xCy = 6C3 = 20 → This is not 35.
Since the only expression that produces 35 when n = 2 and m = 4 is the expression in answer choice B, it is the correct answer.
Answer: B