Question

In how many ways can a committee of 5 members be selected from 6 men and 5 women, consisting 3 men and 2 women?

Answer 1

Hint: Use the fundamental principle of counting to count the number of possible ways of outcome.
In general the number of ways of selecting r people from a group of n people is $^{n}{{C}_{r}}$.
Formula for $^{n}{{C}_{r}}$ is
${{\Rightarrow }^{n}}{{C}_{r}}=\dfrac{n!}{r!\times (n-r)!}$
Complete step-by-step answer:
Total no. of ways of selecting 3 men from 6 men is \[6\mathop{c}_{3}\]
Total no. of ways of selecting 2 women from 5 women is \[5\mathop{c}_{2}\]
By using fundamental principle of counting
Total no. of ways of selecting \[3\]men from \[6\]men and total no. of ways of selecting \[2\] women from women is selected like
\[\Rightarrow 6\mathop{c}_{3}\times 5\mathop{c}_{2}\]
\[\Rightarrow \dfrac{6!}{3!3!}\times \dfrac{5!}{2!3!}\] \[\left( \because n\mathop{c}_{r}=\dfrac{n!}{r!(n-r)!} \right)\]
\[\Rightarrow \dfrac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 3\times 2}\times \dfrac{5\times 4\times 3\times 2\times 1}{2\times 1\times 3\times 2\times 1}\]
\[\Rightarrow 200\]ways
Hence in a committee of 5 members selected from 6 men and 5 women consisting 3 men and 2 women is 200 ways.
Note: The fundamental counting principle is used to count no of possible outcomes.
It explains if there are p ways of doing one event and q ways of doing another event then there are \[p\times q\] ways to perform both of these events.