There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible...

Question

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?

GMAT Panda · Accepted Answer

246 (Option D) 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) = 10!/5! 5! Expand the factorials and cancel 5! in the numerator and denominator. = 10 9 8 7 6 5!5! 5 4 3 2 1 Write the remaining numerator and denominator after cancellation. = 10 9 8 7 6/5 4 3 2 1 Factorize numbers in the numerator and denominator to allow only identical factors to be cancelled. = (5 2) (3 3) (4 2) 7 (3 2)/5 4 3 2 1 Cancel identical factors in the numerator and denominator. = 5 2 3 3 4 2 7 3 25 4 3 2 1 Multiply the remaining numbers to get the total number of ways. = 2 3 2 7 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

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

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