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In how many ways can a committee of 5 persons be formed out of 6 men and 4 women when at least one woman has to be necessarily selected?

seo-qna
Last updated date: 13th Jun 2024
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Answer
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402.9k+ views
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.