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A cosmetics company would like to market a six-item gift

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by swerve » Wed Mar 13, 2019 7:09 am

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A cosmetics company would like to market a six-item gift basket bundle from a set of ten possible items. If 49 of the potential bundles have already been eliminated from consideration, how many potential bundles are still being considered?

A. 147
B. 161
C. 175
D. 182
E. 210

The OA is B

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by GMATGuruNY » Wed Mar 13, 2019 9:33 am
swerve wrote:A cosmetics company would like to market a six-item gift basket bundle from a set of ten possible items. If 49 of the potential bundles have already been eliminated from consideration, how many potential bundles are still being considered?

A. 147
B. 161
C. 175
D. 182
E. 210
From 10 items, the number of ways to choose a bundle of 6 = 10C6 = (10*9*8*7*6*5)/(6*5*4*3*2*1) = 210.
Subtracting the 49 bundles already eliminated from consideration, we get:
210-49 = 161.

The correct answer is B.
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by Scott@TargetTestPrep » Fri Mar 15, 2019 6:29 am
swerve wrote:A cosmetics company would like to market a six-item gift basket bundle from a set of ten possible items. If 49 of the potential bundles have already been eliminated from consideration, how many potential bundles are still being considered?

A. 147
B. 161
C. 175
D. 182
E. 210

The OA is B

Source: Veritas Prep
\

We first can determine how many 6-item bundles can be created from 10 items.

10C6 = 10!/(6!4!) = (10 x 9 x 8 x 7)/(4 x 3 x 2 x 1) = 5 x 3 x 2 x 7 = 210

Since 49 bundles have been removed from consideration, 210 - 49 = 161 bundles are still being considered.

Answer: B

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by [email protected] » Fri Mar 15, 2019 9:53 am
Hi All,

We're told that a cosmetics company would like to market a 6-item gift basket bundle from a set of 10 possible items, but 49 of the potential bundles have already been eliminated from consideration. We're asked for the total number of potential bundles that are still being considered. This question is essentially a 'Combination Formula' question, but there's a great Number Property shortcut that you can use to avoid most of the math...

Choosing 6 items from a group of 10 = 10c6 = (10!)/(6!)(4!).

Since this is a Combination Formula calculation, we know that the end result will be an INTEGER. While that fraction will simplify, we can prove rather quickly that the "10" in the numerator will NOT 'cancel out'.... so the total number of options will be a MULTIPLE OF 10.

When you subtract a number that ends in a '9' (in this case: 49) from a number that ends in a '0'.... you end up with a number that ends in a '1.' There's only one answer that matches...

Final Answer: B

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Contact Rich at [email protected]
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