Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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?

seo-qna
Last updated date: 13th Jun 2024
Total views: 403.2k
Views today: 6.03k
Answer
VerifiedVerified
403.2k+ views
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)!}} $ .

Complete step-by-step answer:
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.