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A committee of $5$ is to be formed out of $6$ gents and $4$ ladies. In how many ways this can be done, if at most two ladies are included?

Last updated date: 13th Jun 2024
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Hint: we find the possible combination of forming a committee. We will add these combinations by considering the ladies in group as zero ladies, one ladies and 2 ladies. We use a formula of combination which is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .

We have to choose a committee of $5$ persons from given that people are $6$ gents and $4$ ladies
The possible way of forming this committee are
There are $5$ gents and $0$ ladies
$4$ gents and $1$ lady
$3$ gents and $2$ ladies
By considering these cases we can find the number of combinations,
The number of combinations of $5$ gents and $0$ ladies is ${}^6{C_5} \times {}^4{C_0}$
Which can further simplified as $\dfrac{{6!}}{{5!\left( {6 - 5} \right)!}} \times \dfrac{{4!}}{{0!\left( {4 - 0} \right)!}}$
Putting the values of factorial then we have,
$= \dfrac{{6!}}{{5!1!}} \times \dfrac{{4!}}{{0!4!}} = 6 \times 1 = 6$
The number of combinations of $4$ gents and $1$ lady is ${}^6{C_4} \times {}^4{C_1}$
Which can further simplified as $\dfrac{{6!}}{{4!\left( {6 - 4} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}$
Putting the values of factorial then we have,
$= \dfrac{{6!}}{{4!2!}} \times \dfrac{{4!}}{{1!3!}}$
Further it can be solved we have,
$= \dfrac{{6 \times 5 \times 4!}}{{4! \times 2}} \times \dfrac{{4 \times 3!}}{{1 \times 3!}} = 15 \times 4 = 60$
The number of combinations of $3$ gents and $2$ ladies is ${}^6{C_3} \times {}^4{C_2}$
Which can further simplified as $\dfrac{{6!}}{{3!\left( {6 - 3} \right)!}} \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}$
Putting the values of factorial then we have,
$= \dfrac{{6!}}{{3!3!}} \times \dfrac{{4!}}{{2!2!}}$
Further it can be solved we have,
$= \dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 6}} \times \dfrac{{4 \times 3 \times 2!}}{{2 \times 2!}} = 20 \times 6 = 120$
Now the total number of combination of forming the committee of $5$ persons is $6 + 60 + 120 = 186$ ways

Note: In calculation we have used a formula to simplify these types of equations is $n! = n \times \left( {n - 1} \right)!$ . Also we should remember the value of some standard factorials which are $0! = 1\& 1! = 1$ . Avoid any type of calculation mistake.