Question

From a group of 4 men and 4 women, 5 persons have to be selected in which at least one man and one woman have to be included. In how many ways can it be done? \[\]
A.320\[\]
B.640\[\]
C.225\[\]
D.56\[\]

Answer 1

Hint: We can select 5 persons such that at least one man and one woman have to be included in either of four cases: 1 man out of 4 men and 4 women out of 4 women, 2 men out of 4 men and 3 women out of 4 women, 3 men out of 4 men and 2 women out of 4 women, 4 men out of 4 men and 1 woman out of 4 women. We find number of ways for each cases using combination ${}^{n}{{C}_{r}}=\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)}{r\left( r-1 \right)...1}$ and add to get the result. \[\]

Complete step by step answer:
We know from combinatorics that the number of ways we can select $r$ objects from $n$ distinct objects is given by
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}=\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)}{r\left( r-1 \right)...1}\]
 We are given in the question that from a group of 4 men and 4 women, 5 persons have to be selected such that at least one man and one woman have to be included. We see that both men and women are distinct objects. We can do it in either of following four ways.\[\]
Case-1: We select 1 man from 4 men in $^{4}{{C}_{1}}$ ways and rest $5-1=4$ women from 4 women in ${}^{4}{{C}_{4}}$ ways. So we use rule of product and find number of ways for case-1 as
\[{{N}_{1}}={}^{4}{{C}_{1}}\times {}^{4}{{C}_{4}}=\dfrac{4}{1}\times \dfrac{4}{4}=4\]
Case-2: We select 2 men from 4 men in $^{4}{{C}_{2}}$ ways and rest $5-2=3$ women from 4 women in ${}^{4}{{C}_{3}}$ ways. So we use rule of product and find number of ways for case-2 as
\[{{N}_{2}}={}^{4}{{C}_{2}}\times {}^{4}{{C}_{3}}=\dfrac{4\times 3}{2\times 1}\times \dfrac{4\times 3\times 2}{3\times 2\times 1}=24\]
Case-3: We select 3 men from 4 men in $^{4}{{C}_{3}}$ ways and rest $5-3=2$ women from 4 women in ${}^{4}{{C}_{2}}$ ways. So we use rule of product and find number of ways for case-3 as
\[{{N}_{3}}={}^{4}{{C}_{3}}\times {}^{4}{{C}_{2}}=\dfrac{4\times 3\times 2}{3\times 2\times 1}\times \dfrac{4\times 3}{1\times 2}=24\]
Case-4: We select 4 men from 4 men in $^{4}{{C}_{4}}$ ways and rest $5-4=1$ woman from 4 women in ${}^{4}{{C}_{1}}$ ways. So we use rule of product and find number of ways for case-4 as
\[{{N}_{4}}={}^{4}{{C}_{4}}\times {}^{4}{{C}_{1}}=\dfrac{4}{4}\times \dfrac{4}{1}=4\]
We use rule of sum and find the total number of ways $N$ to form a group of 5 persons such that one man and one woman have to be included is
\[N={{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}=4+24+24+4=56\]

So, the correct answer is “Option D”.

Note: We note that if there are $m$ ways to do something and $n$ ways to other things then there are $m\times n$ ways to do both things which is called rule of product and $m+n$ ways to both the things which is called rule of sum. If the objects would be identical then we have ${}^{n}{{C}_{r}}=1$. We can group of taking some or all of $n={{n}_{1}}+{{n}_{2}}+...$ things in $\left( {{n}_{1}}+1 \right)\left( {{n}_{2}}+2 \right)...$ ways where ${{n}_{1}}$ things are alike, ${{n}_{2}}$ things are alike and so on.