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Factorials

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Factorials

by gmater29 » Fri Nov 06, 2009 12:47 am
C(m,n) = m! / [(m-n)! n!] for nonnegative integers m and n, m >/ n (m greater then equals 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
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Re: Factorials

by Testluv » Fri Nov 06, 2009 2:32 am
gmater29 wrote:C(m,n) = m! / [(m-n)! n!] for nonnegative integers m and n, m >/ n (m greater then equals 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
Hi gmater,

I'm going to take "x! = 3" as "x does not equal 3" (because, if x is a "nonnegative integer," then it is impossible for x! to equal 3).

We see that this is the combinatorics formula. The equation in the question is telling us what 5C3 equals. C (m,n) and C (5,3) is telling you to sub in 5 for m and 3 for n. And if we use the formula, we find that 5C3 is 10.

The question wants us to figure out what 5CX is. Because 5C3 equals 5CX, and because 5C3 equals 10, we already know that 5CX equals 10.

This question forces us to refer to the answer choices, and the answer chocies are numbers. So we have to backsolve.

Looking at the choices, we can reason that the answer should be closer to 3, and start with either B, C or D.

plugging in, you will find that 5C2 is also 10.

Choose C
Last edited by Testluv on Fri Nov 06, 2009 3:54 am, edited 2 times in total.
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by papgust » Fri Nov 06, 2009 3:21 am
Testluv,

Is it C you are referring to? x must be equal to 2 if 5C2 is correct.
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by Testluv » Fri Nov 06, 2009 3:51 am
thanks papgust; I have edited my post...must be getting sleepy, I thought choice B said "2"!!
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Re: Factorials

by Stuart@KaplanGMAT » Fri Nov 06, 2009 1:13 pm
gmater29 wrote:C(m,n) = m! / [(m-n)! n!] for nonnegative integers m and n, m >/ n (m greater then equals 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
As Testluv states, backsolving is a great way to attack this question.

However, there's another approach we can takes as well.

Once we identify the definition as the combinations formula, there's a rule that, if we know it, we can use to solve:

nCk = nC(n-k)

in which n is the total number of objects available and k is the number that we're choosing.

For example:

10C2 = 10C8

93C21 = 93C72

8C3 = 8C5

3C3 = 3C0

So, once we see that the original gives us 5C3, we automatically know that 5C2 is equivalent.

This rule is particularly helpful for coin flip problems. If we recognize, for example, that the probability of getting 2 head out of 5 flips is the same as that of getting 3 heads out of 5 flips we can often save a lot of time.
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