There are 10 books on a shelf, of which 4 are pape... | Free GMAT Practice Question | GMAT Panda

👋 Hi! I'm GMAT Panda. Ask me anything about this question, or click a button below to get started!

There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

Explanation	Calculations	Help
$To find the number of ways to select 5 books with at least one paperback and one hardback, first count all possible selections, then subtract the cases where all books are of one type.$
$Write the formula for the number of ways to choose 5 books from 10.$	$C (10, 5) = \frac{10 !}{5 ! \times 5 !}$
$Expand the factorials and cancel 5! in the numerator and denominator.$	$= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2 \times 1}$
$Write the remaining numerator and denominator after cancellation.$	$= \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$
$Factorize numbers in the numerator and denominator to allow only identical factors to be cancelled.$	$= \frac{( 5 \times 2 ) \times ( 3 \times 3 ) \times ( 4 \times 2 ) \times 7 \times ( 3 \times 2 )}{5 \times 4 \times 3 \times 2 \times 1}$
$Cancel identical factors in the numerator and denominator.$	$= \frac{5 \times 2 \times 3 \times 3 \times 4 \times 2 \times 7 \times 3 \times 2}{5 \times 4 \times 3 \times 2 \times 1}$
$Multiply the remaining numbers to get the total number of ways.$	$= 2 \times 3 \times 2 \times 7 \times 3 = 252$
$Subtract the selections that are all hardbacks or all paperbacks from the total to get those with at least one of each type.$	$252 - 6 - 0 = 246$

Scroll horizontally to view all columns

246 (Option D)

There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

Explanation	Calculations	Help
$To find the number of ways to select 5 books with at least one paperback and one hardback, first count all possible selections, then subtract the cases where all books are of one type.$
$Write the formula for the number of ways to choose 5 books from 10.$	$C (10, 5) = \frac{10 !}{5 ! \times 5 !}$
$Expand the factorials and cancel 5! in the numerator and denominator.$	$= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2 \times 1}$
$Write the remaining numerator and denominator after cancellation.$	$= \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$
$Factorize numbers in the numerator and denominator to allow only identical factors to be cancelled.$	$= \frac{( 5 \times 2 ) \times ( 3 \times 3 ) \times ( 4 \times 2 ) \times 7 \times ( 3 \times 2 )}{5 \times 4 \times 3 \times 2 \times 1}$
$Cancel identical factors in the numerator and denominator.$	$= \frac{5 \times 2 \times 3 \times 3 \times 4 \times 2 \times 7 \times 3 \times 2}{5 \times 4 \times 3 \times 2 \times 1}$
$Multiply the remaining numbers to get the total number of ways.$	$= 2 \times 3 \times 2 \times 7 \times 3 = 252$
$Subtract the selections that are all hardbacks or all paperbacks from the total to get those with at least one of each type.$	$252 - 6 - 0 = 246$

Scroll horizontally to view all columns

246 (Option D)