Question

Out of $5$ nurses and $2$ doctors, a committee of $3$ members is to be formed. In how many ways can it be done if at least one doctor is to be included?

Answer 1

Whenever we are talking about the combinations (selections) of different objects then we are going to use the concept of combination.
Combination of n different objects taken r at a time is given by $^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .

Complete step by step answer:
As given in the question we have $5$nurses and $2$doctors and we have to form a committee consisting of $3$ members with at least one doctor.
So, we make cases consisting of $3$members as ($1$doctor + $2$ nurses, $2$doctor + 1 nurse).
For selection of persons, we use combinations (i.e., selection of n different objects taken r at a time) $^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
For case $1:$ways to select $1$ doctor and $2$ nurses is $^{2}{{C}_{1}}{{*}^{5}}{{C}_{2}}$ (ways select $1$ doctor with the combination of i.e., $^{2}{{C}_{1}}=$ $\dfrac{2!}{1!\left( 2-1 \right)!}=2$ways and for $2$nurse $^{5}{{C}_{2}}=$ $\dfrac{5!}{2!\left( 5-2 \right)!}=10$ways) which is $2*10=20$ways.
For case $2$: ways to select $2$doctors and $1$nurse is $^{2}{{C}_{2}}{{*}^{5}}{{C}_{1}}$( ways to select $2$doctors $^{2}{{C}_{2}}=\dfrac{2!}{2!\left( 2-2 \right)!}=1$ ways and for nurses $^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}$ $=5$ways.) which is $1*5=5$ ways.
These are the two possibilities so we add both the cases and after that we get the number of ways in which a $3$ member committee is formed with at least one doctor, which is $20+5=25$ ways.

Note: First identify that question is asking about the number of arrangements or combinations then according to the question and try to make possible cases then use the formula to find the case result and lastly add all the results to find the overall result.