Chapter 15.7, Problem 26WE

To determine

To find:

The ways of $4$ student is selected from $6$ girls and $5$ boys.

Answer to Problem 26WE

  $\frac{{}^5{C}_1{\times }^6{C}_3}{{}^5{C}_1{\times }^6{C}_1{\times }^9{C}_2}=\frac{5}{54}$

Explanation of Solution

Given:

The junior and senior class councils each member $=10$ .

Concept used:

Using by the formula of the permutation

  ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

The number of ways of selecting $5$ boys and $6$ girls from $4$ students are selected at random

Such that at least $1$ boy and $1$ girl is selected

  ${=}^5{C}_1{\times }^6{C}_1{\times }^9{C}_2$

  $\begin{array}{l}\frac{{}^5{C}_1{\times }^6{C}_3}{{}^5{C}_1{\times }^6{C}_1{\times }^9{C}_2}=\frac{\frac{5!}{1!\left(5-1\right)!}\times \frac{6!}{3!\left(6-3\right)!}}{\frac{5!}{1!\left(5-1\right)!}\times \frac{6!}{1!\left(6-1\right)!}\times \frac{9!}{2!\left(9-2\right)!}}\\ \frac{{}^5{C}_1{\times }^6{C}_3}{{}^5{C}_1{\times }^6{C}_1{\times }^9{C}_2}=\frac{\frac{5!}{1!4!}\times \frac{6!}{3!3!}}{\frac{5!}{1!4!}\times \frac{6!}{1!5!}\times \frac{9!}{2!7!}}\\ \frac{{}^5{C}_1{\times }^6{C}_3}{{}^5{C}_1{\times }^6{C}_1{\times }^9{C}_2}=\frac{6!}{3!3!}\times \frac{5!}{6!}\times \frac{2!7!}{9!}\\ \frac{{}^5{C}_1{\times }^6{C}_3}{{}^5{C}_1{\times }^6{C}_1{\times }^9{C}_2}=\frac{6\times 5\times 4\times 3!}{3\times 2\times 3!}\times \frac{5!}{6\times 5!}\times \frac{2!7!}{9\times 8\times 7!}\\ \frac{{}^5{C}_1{\times }^6{C}_3}{{}^5{C}_1{\times }^6{C}_1{\times }^9{C}_2}=\frac{5}{54}\end{array}$

Therefore,

  $\frac{{}^5{C}_1{\times }^6{C}_3}{{}^5{C}_1{\times }^6{C}_1{\times }^9{C}_2}=\frac{5}{54}$