Questions & Answers

Question

Answers

(a) 24 ways

(b) 8 ways

(c) 12 ways

(d) 18 ways

Answer

Verified

114.6K+ Views

Hint: For solving this problem we will directly apply one formula of selection of r objects from n objects. Then we will apply the formula in the right manner to get the correct answer.

Complete step-by-step answer:

Given: A bag contains 3 yellow balls and 4 pink balls only and we have to draw 2 pink balls and 1 yellow ball from the bag. We have to find the number of ways in which we can draw 2 pink and 1 yellow ball from this bag.

Now, before we proceed we should one formula that is given below:

The number of ways of selecting $r$ items from the group of $n$ distinct items is:

${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

Now, in this question we select 2 pink balls and 1 yellow ball. We can find the number of ways using the above formulae as follows:

Number of ways of selecting 2 pink balls from 4 pink balls $={}^{4}{{C}_{2}}=\dfrac{4!}{2!\times \left( 4-2 \right)!}=\dfrac{24}{2\times 2!}=6$ .

Number of ways of selecting 1 yellow ball from 3 yellow balls $={}^{3}{{C}_{1}}=\dfrac{3!}{1!\times \left( 3-1 \right)!}=\dfrac{6}{1\times 2!}=3$ .

Now, we have to do both tasks, so for the number of ways in which we can draw 2 pink balls and 1 yellow ball from the bag we have to multiply the above results. Then,

The number of ways to draw 2 pink balls and 1 yellow ball $=6\times 3=18$ ways.

Thus, we can draw 2 pink balls and 1 yellow ball from the bag in 18 ways.

Hence, (d) is the correct option.

Note: Here, the student must take care while simplifying the condition given in the question into different tasks that we have to do and not directly apply the formula to find the total number of ways of selecting 2 pink balls and 1 yellow ball from the bag in the wrong way.

Complete step-by-step answer:

Given: A bag contains 3 yellow balls and 4 pink balls only and we have to draw 2 pink balls and 1 yellow ball from the bag. We have to find the number of ways in which we can draw 2 pink and 1 yellow ball from this bag.

Now, before we proceed we should one formula that is given below:

The number of ways of selecting $r$ items from the group of $n$ distinct items is:

${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

Now, in this question we select 2 pink balls and 1 yellow ball. We can find the number of ways using the above formulae as follows:

Number of ways of selecting 2 pink balls from 4 pink balls $={}^{4}{{C}_{2}}=\dfrac{4!}{2!\times \left( 4-2 \right)!}=\dfrac{24}{2\times 2!}=6$ .

Number of ways of selecting 1 yellow ball from 3 yellow balls $={}^{3}{{C}_{1}}=\dfrac{3!}{1!\times \left( 3-1 \right)!}=\dfrac{6}{1\times 2!}=3$ .

Now, we have to do both tasks, so for the number of ways in which we can draw 2 pink balls and 1 yellow ball from the bag we have to multiply the above results. Then,

The number of ways to draw 2 pink balls and 1 yellow ball $=6\times 3=18$ ways.

Thus, we can draw 2 pink balls and 1 yellow ball from the bag in 18 ways.

Hence, (d) is the correct option.

Note: Here, the student must take care while simplifying the condition given in the question into different tasks that we have to do and not directly apply the formula to find the total number of ways of selecting 2 pink balls and 1 yellow ball from the bag in the wrong way.