Problem Solving

BTGmoderatorDC wrote:10 different biology books and 8 different chemistry books lie on a shelf. In how many ways can a student pick 2 books of each type?

A. 80
B. 160
C. 720
D. 1100
E. 1260

Take the task of selecting 4 books and break it into stages.

Stage 1: Select two biology books
Since the order in which we select the books does not matter, we can use combinations.
We can select 2 books from 10 books in 10C2 ways (45 ways)
So, we can complete stage 1 in 45 ways

If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: https://www.gmatprepnow.com/module/gmat ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

BTGmoderatorDC wrote:10 different biology books and 8 different chemistry books lie on a shelf. In how many ways can a student pick 2 books of each type?

A. 80
B. 160
C. 720
D. 1100
E. 1260

OA E

Source: Economist Gmat

Since the order in which the books are selected does not matter, we have a combination problem. Let's start by determining the number of ways to select the biology books.

Number of ways to select the biology books = 10C2 = 10!/(2! x 8!) = (10 x 9)/2! = 5 x 9 = 45.

Next we can determine the number of ways to select the chemistry books.

Number of ways to select the chemistry books = 8C2 = 8!/(2! x 6!) = (8 x 7)/2! = 4 x 7 = 28.

Thus, the number of ways to select 2 biology books and 2 chemistry books is 45 x 28 = 1,260.

Answer: E