Question

An Engineer Association consists of $5$ civil engineers and $5$ mechanical engineers.

If $2$ civil engineers disagree with each other and refuse to be on the same committee together, how many different ways can a committee of $3$ civil engineers and $2$ mechanical engineers be formed $?$

Answer 1

Number of different ways can a committee of $3$ civil engineers and $2$ mechanical engineers be formed:

Step-1: Find the number of ways to choose $3$ civil engineers from $5$ civil engineers:

Let first and second civil engineers disagree with each other:

Reject first civil engineer and choose from remaining four civil engineers:

$\begin{array}{rcl}{}^{4}C_3\mathrm{ways}& =& \frac{4!}{3!(4-3)!}\\ & =& 4\mathrm{ways}\end{array}$

Reject second civil engineer and choose from remaining four civil engineers:

$\begin{array}{rcl}{}^{4}C_3\mathrm{ways}& =& \frac{4!}{3!(4-3)!}\\ & =& 4\mathrm{ways}\end{array}$

Reject first and second civil engineer and choose from remaining three civil engineers:

$\begin{array}{rcl}{}^{3}C_3\mathrm{ways}& =& \frac{3!}{3!(3-3)!}\\ & =& 1\mathrm{way}\end{array}$

So, total number of ways to choose a committee of $3$ civil engineers and $2$ mechanical engineers when $2$civil engineers disagree with each other will be as follows:

$\begin{array}{rcl}& & ({}^{4}C_3+{}^{4}C_3+{}^{3}C_3)\times \left({}^{5}C_2\right)\end{array}$

$$\begin{gathered} =(4+4+1)\times 10 \\ =90\text{ways}. \end{gathered}$$

Hence, the total number of ways to choose a committee of $3$ civil engineers and $2$ mechanical engineers from a committee of $5$ civil engineers and $5$ mechanical, when $2$civil engineers disagree with each other are $\mathbf{90}\mathbf{}\mathbf{ways}\mathbf{.}$