A gift box contains $5$ chocolates and $7$ candies... | Free GMAT Practice Question | GMAT Panda

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A gift box contains 5 chocolates and 7 candies. If you select 4 treats from the box, how many different selections contain at least one chocolate and at least one candy?

Explanation	Calculations	Help
$We first count all ways to select 4 treats from the box, then subtract the ways that have only chocolates or only candies.$
$We write the number of ways to choose 4 treats from the total of 12 using the Combination Formula .$	$C (12, 4) = \frac{12 !}{4 ! \times 8 !}$	Theory & Tactics Method Card PC3-C Click to view full details
$We expand the factorials and cancel 8! .$	$= \frac{12 \times 11 \times 10 \times 9 \times 8 !}{4 \times 3 \times 2 \times 1 \times 8 !}$
$We factor numbers to allow cancellation of common factors.$	$= \frac{( 4 \times 3 ) \times 11 \times ( 2 \times 5 ) \times 9}{4 \times 3 \times 2 \times 1}$
$We cancel identical factors in the numerator and denominator.$	$= \frac{( 4 \times 3 ) \times 11 \times ( 2 \times 5 ) \times 9}{4 \times 3 \times 2 \times 1}$
$We multiply the remaining numbers to get the total number of selections.$	$= 11 \times 5 \times 9 = 495$
$We find the number of selections that are all chocolates.$	$C (5, 4) = \frac{5 !}{4 ! \times 1 !} = \frac{5 \times 4 !}{4 ! \times 1} = 5$	Theory & Tactics Method Card PC3-C Click to view full details
$We find the number of selections that are all candies.$	$C (7, 4) = \frac{7 !}{4 ! \times 3 !} = \frac{7 \times 6 \times 5 \times 4 \times 3 !}{4 \times 3 \times 2 \times 1 \times 3 !} = 35$	Theory & Tactics Method Card PC3-C Click to view full details
$We subtract the selections with only one type from the total to get the number with at least one of each.$	$495 - 5 - 35 = 455$

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A

A gift box contains 5 chocolates and 7 candies. If you select 4 treats from the box, how many different selections contain at least one chocolate and at least one candy?

Explanation	Calculations	Help
$We first count all ways to select 4 treats from the box, then subtract the ways that have only chocolates or only candies.$
$We write the number of ways to choose 4 treats from the total of 12 using the Combination Formula .$	$C (12, 4) = \frac{12 !}{4 ! \times 8 !}$	Theory & Tactics Method Card PC3-C Click to view full details
$We expand the factorials and cancel 8! .$	$= \frac{12 \times 11 \times 10 \times 9 \times 8 !}{4 \times 3 \times 2 \times 1 \times 8 !}$
$We factor numbers to allow cancellation of common factors.$	$= \frac{( 4 \times 3 ) \times 11 \times ( 2 \times 5 ) \times 9}{4 \times 3 \times 2 \times 1}$
$We cancel identical factors in the numerator and denominator.$	$= \frac{( 4 \times 3 ) \times 11 \times ( 2 \times 5 ) \times 9}{4 \times 3 \times 2 \times 1}$
$We multiply the remaining numbers to get the total number of selections.$	$= 11 \times 5 \times 9 = 495$
$We find the number of selections that are all chocolates.$	$C (5, 4) = \frac{5 !}{4 ! \times 1 !} = \frac{5 \times 4 !}{4 ! \times 1} = 5$	Theory & Tactics Method Card PC3-C Click to view full details
$We find the number of selections that are all candies.$	$C (7, 4) = \frac{7 !}{4 ! \times 3 !} = \frac{7 \times 6 \times 5 \times 4 \times 3 !}{4 \times 3 \times 2 \times 1 \times 3 !} = 35$	Theory & Tactics Method Card PC3-C Click to view full details
$We subtract the selections with only one type from the total to get the number with at least one of each.$	$495 - 5 - 35 = 455$

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A