C(m, n) = m!/((m - n)!n!) for nonnegatives integers m and n, m n.

Question

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? (A) 0 (B) 1 (C) 2 (D) 4 (E) 5 2024-01-24_19-55-50.png

KarishmaB · Accepted Answer

The given expression should remind you that it is mCn i.e. selecting n out of total m items.

We know that mCn = mC(m-n) and hence 5C3 = 5C(5-3) = 5C2

Even if this doesn't strike you during the exam, simply substitute.

(m!)/((m-n)!n!) = 5!/3!*2! = 5!/(5-x)!*x!

(5-x)!*x! = 3!*2!

We see that x = 2

Answer (C)

acegmat123 · Answer

In Combinatorics,

nCr = nC(n-r)

So x = 5-3 =2

C

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Kinshook · Answer

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

MartyMurray · Answer

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:  C

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C(m, n) = m!/((m - n)!n!) for nonnegatives integers m and n, m n.

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C(m, n) = m!/((m - n)!n!) for nonnegatives integers m and n, m n.

Prep Toolkit

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