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Permutation and Combination Problem

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by sukhman » Mon Sep 09, 2013 10:35 pm
In a college, 8 students play at the State level and 10 at the National level. If 6 students play at both National and State levels, in how many ways can 9 students be selected from among these? 12C9
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Source: — Problem Solving |

by ganeshrkamath » Mon Sep 09, 2013 11:59 pm
sukhman wrote:In a college, 8 students play at the State level and 10 at the National level. If 6 students play at both National and State levels, in how many ways can 9 students be selected from among these? 12C9
6 play at both levels
(8-6) = 2 play at only state level
(10-6) = 4 play at only national level
Total = 6 + 2 + 4 = 12
Select 9 out of 12 = 12C9

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by Brent@GMATPrepNow » Tue Sep 10, 2013 6:49 am
ganeshrkamath's solution is perfect, and the answer is, indeed, 12C9
However, it's unlikely that the answer choices would be in the form xCy (since there are different ways to denote combinations). For this particular question, we'd probably have to evaluate 12C9. So, let's do that (quickly!!).

If anyone is interested, we have a free video on calculating combinations in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

When calculating combinations (in our head), it's useful to know the following rule: nCr = nC(n-r). In other words, "n choose r" is equal to "n choose n-r"
So, for example: 10C7 = 10C3
8C7 = 8C1
20C18 = 20C2

This rule is useful because it's much easier to calculate combinations when the second value is smaller.

So, here 12C9 = 12C3
= (12)(11)(10)/(3)(2)(1)
= (2)(11)(10)
= 220

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Brent
Brent Hanneson - Creator of GMATPrepNow.com
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