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C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n. If C(5,3)=C(5,x) and x≠3, what is the value of x?


(A) 0
(B) 1
(C) 2
(D) 4
(E) 5

Given: C(m,n) = (m!)/((m-n)!n!) for nonnegatives integers m and n, m>=n.
Asked: If C(5,3)=C(5,x) and x≠3, what is the value of x?

C(5,3) = 5!/(3!2!)
C(5,2) = 5!/(2!3!)

IMO C
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\(C(m,n) = \frac{m!}{(m-n)!n!}\) for nonnegatives integers m and n, m ≥ n. If C(5, 3) = C(5, x) and x≠3, what is the value of x?

Looking at \(C(m,n) = \frac{m!}{(m-n)!n!}\), we can recognize that it's the combinations formula for choosing n out of m elements.

So, C(5, 3) is the same as 5c3.

Now, a key fact about combinations is that the number of combinations of XcY is the same as the number of combinations of Xc(X - Y).

The reason why is that, when we choose Y elements from a group of X elements, we are also leaving X - Y elements out of X elements.

In other words, every combination of Y elements out of X is paired with a combination of X - Y elements out of X.

For example, each time we choose a combination of 3 elements out of 5, we also leave a combination of 5 - 3 = 2 elements out of 5.

So, since there are 10 combinations of 3 elements out of 5, there are also 10 combinations of 2 elements out of 5.

Thus, C(5, 3) = C(5, 2).

(A) 0
(B) 1
(C) 2
(D) 4
(E) 5

­
Correct answer:
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It is a simple known property of combinations that you can memorize. It would come handy in exam:
nCr = nC(n-r)
so,
5C3= 5C(5-3) = 5C2
Hence, X= 2
answer C
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