Question

How many committees of $5$ members each can be formed from $8$ official and $4$ non-official members in the following cases-
a.Each consisting of $3$ official and $2$ non-official members.
b.Each contains at least two non-official members.
c.A particular official member is never included.
d.A particular non-official member is always included.

Answer 1

To select r members out of n total members, we use the following formula-${}^{\text{n}}{{\text{C}}_{\text{r}}}$
And we know that-
Formula of combination is given as-
$\Rightarrow$ ${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\dfrac{{n!}}{{r!n - r!}}$Where n=total number of things and r=no. of things to be selected
Use these two formulas to get the answer.

Complete step-by-step answer:
Given total number of official members = $8$
Total number of non-official members = $4$
The total number of committees to be formed = $5$
(a) We have to form such committees that consist of $3$ official and $2$ non-official members.
So the number of ways to select $3$ official members out of $8$ official members = ${}^8{C_3}$
On applying formula of combination which is given as-
$\Rightarrow$ ${}^{\text{n}}{{\text{C}}_{\text{r}}}$=$\dfrac{{n!}}{{r!n - r!}}$Where n=total number of things and r=no. of things to be selected
We have-
The number of ways to select $3$ official members out of $8$ official members = $\dfrac{{8!}}{{3!8 - 3!}} = \dfrac{{8 \times 7 \times 6 \times 5!}}{{3 \times 2 \times 5!}}$
On solving we get,
The number of ways to select $3$ official members out of $8$ official members =$8 \times 7 = 56$ ways
The number of ways to select $2$ non-official members from $4$ non-official members = ${}^4{C_2}$
On applying formula of combination we get,
The number of ways to select $2$ non-official members from $4$ non-official members = $\dfrac{{4!}}{{2!4 - 2!}} = \dfrac{{4 \times 3 \times 2}}{4} = 6$ ways
Then the total number of ways committees consists of $3$ official and $2$ non-official members=$56 \times 6 = 336$ ways
The total number of ways committees consists of $3$ official and $2$ non-official members=$336$ ways
(b) We have to find the number of ways that each committee contains at least two non-official members.
So the number of ways that each committee contains at least two non-official members= total number of ways five members can be selected from $12$ members - total number of ways the committees contains one non-official member- total number of ways the committees contain no non-official number
So we can write is as-
The number of ways that each committee contains at least two non-official members=${}^{12}{C_5} - \left( {{}^8{C_4} \times {}^4{C_1}} \right) - {}^8{C_5}$
On applying the formula of combination we get,
The number of ways that each committee contains at least two non-official members= $\dfrac{{12!}}{{5!9!}} - \left( {\dfrac{{8!}}{{4!4!}} \times 4} \right) - 56$
On solving we get,
The number of ways that each committee contains at least two non-official members=$792 - 280 - 56 = 456$
The number of ways that each committee contains at least two non-official members= $456$ ways.
(c)We have to find the number of ways a particular official member is never included.
So excluding the particular official member the total number of members is $11$
Then the number of ways a particular official member is never included= ${}^{11}{C_5} = \dfrac{{11!}}{{6!5!}}$
On solving we get,
The number of ways a particular official member is never included= $\dfrac{{11 \times 10 \times 9 \times 8 \times 7}}{{5 \times 4 \times 3 \times 2}} = 11 \times 3 \times 4 \times 7 = 462$
The number of ways a particular official member is never included = $462$ ways.
(d)We have to find the total number of ways a particular member is always included in the committees.
Since one particular member is always included then we only have find the number of ways to form $4$ member committees from $11$ members
Total number of ways a particular member is always included in the committees= ${}^{11}{C_4} = \dfrac{{11!}}{{7!4!}}$
On solving we get,
Total number of ways a particular member is always included in the committees=$\dfrac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2}} = 3 \times 110 = 330$
Total number of ways a particular member is always included in the committees=$330$ ways.

Note: In this question we have used combinations because we have to form committees only which can be formed in any order. Combination is used when things are to be arranged but not necessarily in order. Permutation is a little different. In permutation, order is important. Permutation is given by-
$\Rightarrow {}^n{P_r} = \dfrac{{n!}}{{n - r!}}$ Where n=total number of things and r=no. of things to be selected.