A collection contains $11$ DVDs, of which $5$ are ... | Free GMAT Practice Question | GMAT Panda

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Question

A collection contains 11 DVDs, of which 5 are action films and 6 are romance films. How many ways are there to select 4 DVDs so that the selection includes at least one action film and at least one romance film?

Answer Choices

A. $310$
B. $315$
C. $320$
D. $330$
E. $300$

Steps

Explanation	Calculations	Help
$We apply Combinatorics Inverse Rule : we count all possible selections of 4 DVDs, then subtract those with no action films and those with no romance films.$	$Valid ways = Total ways - No action ways - No romance ways$	Theory & Tactics Method Card PC2-C Click to view full details
$We use Combination Formula to find the number of ways to choose any 4 DVDs from 11.$	$\frac{11 !}{4 ! \times 7 !}$	Theory & Tactics Method Card PC3-C Click to view full details
$We expand the factorials and cancel the common 7! factor.$	$= \frac{11 \times 10 \times 9 \times 8 \times 7 !}{( 4 \times 3 \times 2 \times 1 ) \times 7 !}$	Theory & Tactics Method Card CALC2-C Click to view full details
$We write the remaining fraction after cancellation.$	$= \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}$	Theory & Tactics Method Card CALC2-C Click to view full details
$We break down the numerator factors to identify common factors with the denominator.$	$= \frac{11 \times ( 5 \times 2 ) \times ( 3 \times 3 ) \times ( 4 \times 2 )}{4 \times 3 \times 2 \times 1}$	Theory & Tactics Method Card CALC2-C Click to view full details
$We cancel the common factors.$	$= \frac{11 \times 5 \times 2 \times 3 \times 3 \times 4 \times 2}{4 \times 3 \times 2 \times 1}$	Theory & Tactics Method Card CALC2-C Click to view full details
$We multiply the remaining factors to get the total number of ways.$	$= 11 \times 5 \times 3 \times 2 = 330$
$We use Combination Formula to find the number of selections with no romance films (all are action films).$	$\frac{5 !}{4 ! \times 1 !} = \frac{5 \times 4 !}{4 ! \times 1 !} = 5$	Theory & Tactics Method Card PC3-C Click to view full details
$We use Combination Formula to find the number of selections with no action films (all are romance films).$	$\frac{6 !}{4 ! \times 2 !} = \frac{6 \times 5 \times 4 !}{4 ! \times ( 2 \times 1 )} = \frac{6 \times 5}{2 \times 1} = 15$	Theory & Tactics Method Card PC3-C Click to view full details
$We subtract the undesired cases from the total to get the number of selections with at least one of each genre.$	$330 - 5 - 15 = 310$

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Final Answer

A

Question

A collection contains 11 DVDs, of which 5 are action films and 6 are romance films. How many ways are there to select 4 DVDs so that the selection includes at least one action film and at least one romance film?

Answer Choices

A. $310$
B. $315$
C. $320$
D. $330$
E. $300$

Steps

Explanation	Calculations	Help
$We apply Combinatorics Inverse Rule : we count all possible selections of 4 DVDs, then subtract those with no action films and those with no romance films.$	$Valid ways = Total ways - No action ways - No romance ways$	Theory & Tactics Method Card PC2-C Click to view full details
$We use Combination Formula to find the number of ways to choose any 4 DVDs from 11.$	$\frac{11 !}{4 ! \times 7 !}$	Theory & Tactics Method Card PC3-C Click to view full details
$We expand the factorials and cancel the common 7! factor.$	$= \frac{11 \times 10 \times 9 \times 8 \times 7 !}{( 4 \times 3 \times 2 \times 1 ) \times 7 !}$	Theory & Tactics Method Card CALC2-C Click to view full details
$We write the remaining fraction after cancellation.$	$= \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}$	Theory & Tactics Method Card CALC2-C Click to view full details
$We break down the numerator factors to identify common factors with the denominator.$	$= \frac{11 \times ( 5 \times 2 ) \times ( 3 \times 3 ) \times ( 4 \times 2 )}{4 \times 3 \times 2 \times 1}$	Theory & Tactics Method Card CALC2-C Click to view full details
$We cancel the common factors.$	$= \frac{11 \times 5 \times 2 \times 3 \times 3 \times 4 \times 2}{4 \times 3 \times 2 \times 1}$	Theory & Tactics Method Card CALC2-C Click to view full details
$We multiply the remaining factors to get the total number of ways.$	$= 11 \times 5 \times 3 \times 2 = 330$
$We use Combination Formula to find the number of selections with no romance films (all are action films).$	$\frac{5 !}{4 ! \times 1 !} = \frac{5 \times 4 !}{4 ! \times 1 !} = 5$	Theory & Tactics Method Card PC3-C Click to view full details
$We use Combination Formula to find the number of selections with no action films (all are romance films).$	$\frac{6 !}{4 ! \times 2 !} = \frac{6 \times 5 \times 4 !}{4 ! \times ( 2 \times 1 )} = \frac{6 \times 5}{2 \times 1} = 15$	Theory & Tactics Method Card PC3-C Click to view full details
$We subtract the undesired cases from the total to get the number of selections with at least one of each genre.$	$330 - 5 - 15 = 310$

Scroll horizontally to view all columns

Final Answer

A