Question

Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is
A. 200
B. 300
C. 500
D. 350

Answer 1

In this question, we have to select the team, so, we will use combination theory. First we will find the total number of ways of selecting 2 girls and 3 boys. After this, we will calculate the ways in which A and B are always included in the same team and finally subtract them to get the answer.

Complete step-by-step answer:
We have 5 girls and 7 boys and our team should consist of 2 girls and 3 boys such that A and B are members of the same team.
So first of all we will calculate the total ways of forming a team consisting of 2 girls and 3 boys.
Total number of ways = ${}^5{C_2}.{}^7{C_3}$
we know that ${}^n{C_r}$is also calculated as, ${}^n{C_r} = \dfrac{{n \times (n - 1)(n - 2)....{\text{to r factors}}}}{{r!}}$
For example - ${}^4{C_2} = \dfrac{{4 \times 3}}{{2!}} = \dfrac{{4 \times 3}}{{2 \times 1}} = 6$
So, we can write:
Total number of ways = ${}^5{C_2}.{}^7{C_3} = \dfrac{5}{2} \times \dfrac{4}{1} \times \dfrac{7}{3} \times \dfrac{6}{2} \times \dfrac{5}{1} = 350$
Now if we subtract the ways in which A and B are always included in the same team from this then we will get our answer.
So when A and B are both included then we take only 1 boy from the remaining 5 boys and 2 girls from 5 girls.
$\therefore$ Number of ways of when A and B are always included = ${}^5{C_1}.{}^5{C_2} = \dfrac{5}{1} \times \dfrac{5}{2} \times \dfrac{4}{1} = 50$.
Therefore, require number of ways = Total number of ways - Number of ways of when A and B are always included = 350 – 50 = 300.

So, the correct answer is “Option B”.

Note: In this type question, it is easy to calculate the total ways and subtract the conditions which are not required. You can also calculate the combination as, ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$, where ${}^n{C_r}$ denotes the number of ways of combinations of n different things taken r at a time. In case of selection, we use combination and in case of selection, we use permutation.