A bakery has $5$ chocolate muffins and $7$ vanilla... | Free GMAT Practice Question | GMAT Panda

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A bakery has 5 chocolate muffins and 7 vanilla muffins. In how many ways can you select 4 muffins from these if your selection must include at least one chocolate muffin and at least one vanilla muffin?

Explanation	Calculations	Help
$To find the number of ways to select 4 muffins with at least one chocolate and at least one vanilla, we first count all possible selections of 4 muffins from the total, then subtract the cases where all muffins are of one flavor.$
$Write the formula for the total number of ways to choose 4 muffins from the 12 available.$	$C (12, 4) = \frac{12 !}{4 ! \times 8 !}$	Theory & Tactics Method Card PC3-C Click to view full details
$Expand the factorials and cancel 8! in the numerator and denominator.$	$= \frac{12 \times 11 \times 10 \times 9 \times 8 !}{4 ! \times 8 !}$	Theory & Tactics Method Card CALC2-C Click to view full details
$Write the remaining numerator and denominator after cancellation.$	$= \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$
$Factorize numbers in the numerator and denominator to allow identical factors to be cancelled.$	$= \frac{( 4 \times 3 ) \times 11 \times ( 2 \times 5 ) \times ( 3 \times 3 )}{4 \times 3 \times 2 \times 1}$
$Cancel identical factors in the numerator and denominator.$	$= \frac{4 \times 3 \times 11 \times 2 \times 5 \times 3 \times 3}{4 \times 3 \times 2 \times 1}$
$Multiply the remaining numbers to get the total number of ways.$	$= 11 \times 5 \times 3 \times 3 = 495$
$Subtract the selections that are all chocolate or all vanilla to get those with at least one of each flavor.$	$495 - C (5, 4) - C (7, 4) = 495 - 5 - 35 = 455$	Theory & Tactics Method Card PC2-C Click to view full details

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B

A bakery has 5 chocolate muffins and 7 vanilla muffins. In how many ways can you select 4 muffins from these if your selection must include at least one chocolate muffin and at least one vanilla muffin?

Explanation	Calculations	Help
$To find the number of ways to select 4 muffins with at least one chocolate and at least one vanilla, we first count all possible selections of 4 muffins from the total, then subtract the cases where all muffins are of one flavor.$
$Write the formula for the total number of ways to choose 4 muffins from the 12 available.$	$C (12, 4) = \frac{12 !}{4 ! \times 8 !}$	Theory & Tactics Method Card PC3-C Click to view full details
$Expand the factorials and cancel 8! in the numerator and denominator.$	$= \frac{12 \times 11 \times 10 \times 9 \times 8 !}{4 ! \times 8 !}$	Theory & Tactics Method Card CALC2-C Click to view full details
$Write the remaining numerator and denominator after cancellation.$	$= \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$
$Factorize numbers in the numerator and denominator to allow identical factors to be cancelled.$	$= \frac{( 4 \times 3 ) \times 11 \times ( 2 \times 5 ) \times ( 3 \times 3 )}{4 \times 3 \times 2 \times 1}$
$Cancel identical factors in the numerator and denominator.$	$= \frac{4 \times 3 \times 11 \times 2 \times 5 \times 3 \times 3}{4 \times 3 \times 2 \times 1}$
$Multiply the remaining numbers to get the total number of ways.$	$= 11 \times 5 \times 3 \times 3 = 495$
$Subtract the selections that are all chocolate or all vanilla to get those with at least one of each flavor.$	$495 - C (5, 4) - C (7, 4) = 495 - 5 - 35 = 455$	Theory & Tactics Method Card PC2-C Click to view full details

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B