Question

A committee of $4$ persons is to be formed from $2$ ladies, $2$ old men and $4$ young men such that it includes at least $1$ lady, at least $1$ old man and at most $2$ young men. Then the total number of ways in which this committee can be formed is

Answer 1

We will first find out all the possible cases which are possible and then we will find the combinations from them. Finally we will add all possibilities to get the answer.

Formula used: $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$

Complete step-by-step solution:
From the question it is given that the committee should comprise $4$ people.
There are a total $2$ ladies, $2$ old men and $4$ young men from which the committee has to be formed.
Now, there has to be at least $1$ lady in the committee therefore the number of ladies which can be in the committee is $1$ or $2$.
Now, there has to be at least $1$ old man in the committee therefore the number of old men who can be in the committee is $1$ or $2$.
Now, there can be at most $2$ young men in the committee therefore the total number of young men shouldn’t exceed $2$, since there is no at least value to the participation of young men, there can be $0$,$1$ or $2$ men in the committee.
The first case is: $1$ lady, $1$ old man and $2$ young.
The total number of ways in which this can be done is: $^2{C_1}{ \times ^2}{C_1}{ \times ^4}{C_2}$
On splitting this combination by using the formula and we get
$\Rightarrow 2 \times 2 \times \dfrac{{4 \times 3}}{{2 \times 1}}$
On further simplification we get:
$\Rightarrow 24$
Now, the second case is: $2$ ladies, $1$ old man and $1$ young.
The total number of ways in which this can be done is:
$^2{C_2}{ \times ^2}{C_1}{ \times ^4}{C_1}$
On splitting this combination by using the formula and we get
$\Rightarrow 1 \times 1 \times 4$
On further simplification we get:
$\Rightarrow 8$.
Now, the third case is: $2$ ladies,$2$ old men and $0$ young.
The total number of ways in which this can be done is:
$^2{C_2}{ \times ^2}{C_2}$
On splitting this combination by using the formula and we get:
$\Rightarrow 1 \times 1$
On further simplification we get:
$\Rightarrow 1$.
Now, the fourth case is: $1$ ladies, $2$ old men and $1$ young.
The total number of ways in which this can be done is:
$^2{C_1}{ \times ^2}{C_2}{ \times ^4}{C_1}$
On splitting this combination by using the formula and we get
$\Rightarrow 2 \times 1 \times 4$
On further simplification we get:
$\Rightarrow 8$.
Now the total ways in which they could be selected is all the cases added therefore, it can be written as:
$\Rightarrow 24 + 8 + 1 + 8$
On adding the terms
$\Rightarrow 41$.

Therefore the total number of ways in which a committee can be formed is 41.

Note: The combination formula was used in this question, it finds the subset of items, in combination, and the order does not matter.
Also it is exists a permutation formula which is: $^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$ Which finds number of ways elements can be arranged, here the order matters.