Answer

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**Hint:**Here we are using the given condition (at least one lady should be there) we will have four possible ways to form a committee with at least one lady.

Here we use the combination formula and then we will add them.

**Formula used:**

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

**Complete step by step answer:**

It is given that there are $6$ Men and $4$ women.

We have to find out how many ways we can make a committee of $5$ persons from $6$ men and $4$ women with at least $1$ women.

According to the question we need to make a committee of $5$ and each committee formed there must be a lady. There is $6$ men and $4$ women at least one woman in the following four ways as done below:

Now we have to form in given committee as the following ways:

($1$ Lady $ + 4$ Men) or ($2$ women$ + 3$ men) or ($3$ women $ + 2$ men) or ($4$ women $ + 1$ man) or ($5$ women$ + 0$ gents)

Hence the total number of possible arrangements:

$ = ({}^4{C_1} \times {}^6{C_4}) + ({}^4{C_2} \times {}^6{C_3}) + ({}^4{C_3} \times {}^6{C_2}) + ({}^4{C_4} \times {}^6{C_1})....\left( 1 \right)$

Here we have to split the term and we find one by one we get,

First we take $({}^4{C_1} \times {}^6{C_4})$

Here we use the formula for combination ${}^n{C_r} = \dfrac{{n!}}{{r!(n - 1)!}}$ and we get,

$\Rightarrow ({}^4{C_1} \times {}^6{C_4}) = \dfrac{{4!}}{{1!(4 - 1)!}} \times \dfrac{{6!}}{{4!(6 - 4)!}}$

On subtracting the denominator term we get,

$ \Rightarrow \dfrac{{4!}}{{1!(3)!}} \times \dfrac{{6!}}{{4!(2)!}}$

On splitting the factorial term we get,

$ \Rightarrow \dfrac{{4 \times 3!}}{{1!(3)!}} \times \dfrac{{6 \times 5 \times 4!}}{{4!(2)!}}$

On cancelling the same term we get,

$ \Rightarrow \dfrac{4}{1} \times \dfrac{{6 \times 5}}{{2 \times 1}}$

On dividing the term we get,

$ \Rightarrow 4 \times 3 \times 5$

Let us multiplying the term we get,

$ \Rightarrow 60$

Also we take the second term, $({}^4{C_2} \times {}^6{C_3})$

Here we use the formula for combination ${}^n{C_r} = \dfrac{{n!}}{{r!(n - 1)!}}$ and we get,

$\Rightarrow ({}^4{C_2} \times {}^6{C_3}) = \dfrac{{4!}}{{2!(4 - 2)!}} \times \dfrac{{6!}}{{3!(6 - 3)!}}$

On subtracting the denominator term we get,

$ \Rightarrow \dfrac{{4!}}{{2!(2)!}} \times \dfrac{{6!}}{{3!(3)!}}$

On splitting the factorial term we get,

$ \Rightarrow \dfrac{{4 \times 3 \times 2!}}{{2!(2)!}} \times \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!(3)!}}$

On cancelling the same term we get,

$ \Rightarrow \dfrac{{4 \times 3}}{{2 \times 1}} \times \dfrac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}}$

On dividing the term we get,

$ \Rightarrow 2 \times 3 \times 5 \times 4$

Let us multiplying the term we get,

$ \Rightarrow 120$

Also we take the third term, $({}^4{C_3} \times {}^6{C_2})$

Here we use the formula for combination ${}^n{C_r} = \dfrac{{n!}}{{r!(n - 1)!}}$ and we get,

$\Rightarrow ({}^4{C_3} \times {}^6{C_2}) = \dfrac{{4!}}{{3!(4 - 3)!}} \times \dfrac{{6!}}{{2!(6 - 2)!}}$

On subtracting the denominator term we get,

$ \Rightarrow \dfrac{{4!}}{{3!(1)!}} \times \dfrac{{6!}}{{2!(4)!}}$

On splitting the factorial term we get,

$ \Rightarrow \dfrac{{4 \times 3!}}{{3!(1)!}} \times \dfrac{{6 \times 5 \times 4!}}{{2!(4)!}}$

On cancelling the same term we get,

$ \Rightarrow \dfrac{4}{1} \times \dfrac{{6 \times 5}}{{2 \times 1}}$

On dividing the term we get,

$ \Rightarrow 4 \times 3 \times 5$

Let us multiplying the term we get,

$ \Rightarrow 60$

Finally we take that, $({}^4{C_4} \times {}^6{C_1})$

Here we use the formula for combination ${}^n{C_r} = \dfrac{{n!}}{{r!(n - 1)!}}$ and we get,

$({}^4{C_4} \times {}^6{C_1}) = \dfrac{{4!}}{{4!(4 - 4)!}} \times \dfrac{{6!}}{{1!(6 - 1)!}}$

On subtracting the denominator term we get,

$ \Rightarrow \dfrac{{4!}}{{4!(0)!}} \times \dfrac{{6!}}{{1!(5)!}}$

On splitting the factorial term we get,

$ \Rightarrow \dfrac{{4!}}{{4! \times 1}} \times \dfrac{{6 \times 5!}}{{1!(5)!}}$

On cancelling the same term we get,

$ \Rightarrow 6$

On substituting the value in \[\left( 1 \right)\] we get,

$ = 60 + 120 + 60 + 6$

By adding the terms

$ = 246$

**$\therefore$The total number of ways that the committee can be formed with the given condition = 246**

**Note:**

Whenever you face such types of problems you have to think of a number of ways.

Then according to the condition provided you have to select people with the help of combination as it is used for selections. ${}^n{C_r}$ By using this we get the answer.

This way will give you the right answer.

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