Answer
Verified
439.8k+ 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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE