Question

Josephine has five chemistry books, three history books, and six statistics books. She wants to choose one book of each type to study. In how many ways can she choose the three books?

Answer 1

Answer:

150

Step-by-step explanation:

number of Chemistry books = 5

number of History books =  3

number of Statistics books = 6

Number of ways to select one Chemistry book out of 5

= $^{5}C_{3}=\frac{5!}{3!\times 2!}=10$

Number of ways to select one History book out of 3

= $^{3}C_{1}=\frac{3!}{1!\times 2!}=3$

Number of ways to select one Statistics book out of 6

= $^{6}C_{1}=\frac{6!}{1!\times 5!}=5$

So, the total number of ways to select three books = 10 x 3 x 5 = 150

Thus, there are 150 ways to select three books.  

Answer 2

Using the Fundamental Counting Theorem, it is found that she can choose the three books in 90 ways.

Fundamental counting theorem:

States that if there are n things, each with $n_1, n_2, …, n_n$ ways to be done, each thing independent of the other, the number of ways they can be done is:

$N = n_1 \times n_2 \times \cdots \times n_n$

In this problem:

• 5 chemistry books, hence $n_1 = 5$
• 3 history books, hence $n_2 = 3$
• 6 statistics books, hence $n_3 = 6$

Then:

$N = 5 \times 3 \times 6 = 90$

She can choose the three books in 90 ways.

A similar problem is given at https://brainly.com/question/19022577