# a committee of 5 persons is to be formed from 6 men and 4 women. in how many ways can this be done when at least 3 women are included?

### Mohammed

Guys, does anyone know the answer?

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## A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways can this be done, when (i) at least 2 ladies are included? (ii) at most 2 ladies are included?

Answer : Since the committee of 5 is to be formed from 6 gents and 4 ladies. (i) Forming a committee with at least 2 ladies Here the possibilities are 2 ladies and 3 gents 3 ladies and 2 gents 4 ladies and 1 gent The number of ways they can be selected = 186 […]

A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways can this be done, when (i) at least 2 ladies are included? (ii) at most 2 ladies are included?

Rs aggarwal Maths CBSE Class 11 Chapter 9 Exercise 9B

## A committee of 5 is to be formed out of 6 men and 4 ladies. In how many ways can this be done, when (a) at least 2 ladies are included, (b) at most 2 ladies are included?

(a) We have to make a selection of (i) (2 ladies out of 4) and (3 men out of 6) or (ii) (3 ladies out of 4 ) and (2 men out of 6) or (iii) (4 ladies out of 4) and (1 man out of 6). The number of ways of these selections are: Case I .^(4)C(2)xx.^(6)C(3)=6xx20=120. Case II .^(4)C(3)xx.^(6)C(2)=4xx15=60. Case III .^(4)C(4)xx.^(6)C(1)=1xx6=6. Hence, the required number of ways =(120+60+6)=186. (b) We have to make a selection of (i) (1 lady out of 4) and (4 men out of 6) or (ii) (2 ladies out of 4) and (3 men out of 6). The number of ways of these selections are: Case I .^(4)C(1)xx.^(6)C(4)=4xx15=60. CaseII .^(4)C(2)xx.^(6)C(4)=4xx15=60. Case II .^(4)C(2)xx.^(6)C(3)=6xx20=120. Hence, the required number of ways = (60+120)=180.

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A committee of 5 is to be form...

A committee of 5 is to be formed out of 6 men and 4 ladies. In how many ways can this be done, when

(a) at least 2 ladies are included,

(b) at most 2 ladies are included?

Updated On: 27-06-2022

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(a) We have to make a selection of

(i) (2 ladies out of 4) and (3 men out of 6)

or (ii) (3 ladies out of 4 ) and (2 men out of 6)

or (iii) (4 ladies out of 4) and (1 man out of 6).

The number of ways of these selections are:

Case I . 4 C 2 × . 6 C 3 =6×20=120 .4C2×.6C3=6×20=120 . Case II . 4 C 3 × . 6 C 2 =4×15=60 .4C3×.6C2=4×15=60 . Case III . 4 C 4 × . 6 C 1 =1×6=6 .4C4×.6C1=1×6=6 .

Hence, the required number of ways =(120+60+6)=186.

(b) We have to make a selection of

(i) (1 lady out of 4) and (4 men out of 6)

or (ii) (2 ladies out of 4) and (3 men out of 6).

The number of ways of these selections are:

Case I . 4 C 1 × . 6 C 4 =4×15=60 .4C1×.6C4=4×15=60 . CaseII . 4 C 2 × . 6 C 4 =4×15=60 .4C2×.6C4=4×15=60 . Case II . 4 C 2 × . 6 C 3 =6×20=120 .4C2×.6C3=6×20=120 .

Hence, the required number of ways = (60+120)=180.

Answer

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## A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included?

Answer (1 of 20): The number of ways is 186, as noted in other answers. Here’s another way to do it: take all possible ways of choosing five people, and subtract the ways that are all five men or four men and one woman: {10 \choose 5} - {6 \choose 5}{4 \choose 0} - {6 \choose 4}{4 \choose 1} = 18...

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included?

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Sort Abhinav

Studied at Indian Institute of Technology Varanasi (IIT BHU) (Graduated 2018)5y

Let us assume that the available men be denoted by M1, M2, M3, M4, M5 and M6 and available women be denoted by W1, W2, W3 and W4. Let’s assume that the chosen women are W1 and W2 (by 4C2) and other members be M1, M2 and W3 (by 8C3). So, the committee contains members M1, M2, W1, W2 and W3. Now, let’s assume that the chosen women be W1 and W3 (by 4C2) and other members be M1, M2 and W2 (by 8C3). Now, the committee contains the members M1, M2, W1, W2 and W3 (same as before) but it is counted as 2 different cases in 4C2*8C3. So, the number of ways given by 4C2*8C3 contains repeated cases. The num

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Jeremy van der Smeede

Studies Applied Physics at Delft University of Technology (Expected 2024)3y

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In how many ways can a committee of 5 members be formed out of 6 boys and 4 girls, if at least 2 girls are in the committee?

Alright so let’s see what we have here. There are 5 possible members 2 of them are girls. This means that for the those 2 seats we have 4 and then 3 possible individuals that can be seeted there! The last 3 seats can be occupied by first 1 out 8 individuals then 7 individuals and then 6 individuals. This means that there is a chance of 4*3*8*7*6= 4032 possible ways to form this committee

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Shreya 5y

You have to make cases for this question.

Total men available=6

Total women available =4

Condition atleast two women.

Case I: 2 women 3 men

This can be done in 4c2 x 6c3

Case II: 3 women 2 men 4c3 x 6c2 Case III: 4 women 1 men 4c4 x 6c1

This question can also be done by multinomial theorm also.

Vikas Singh

Have studied mathematics in school and college5y

Here it says atleast 2 women, which in other words mean, the committee may comprise of 2 women and 3 men or 3 women and 2 men or 4 women and 1 man.

Therefore the answer should be 4C2*6C3 + 4C3*6C2 + 4C4*6C1 = 6*20 + 4*15 + 1*6 = 186

Now lets see why the answer should not be 4C2*8C3 = 6*56 = 336

The two groups of items to select from are not identical so you can't combine them together to select from the combined list.

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A committee of 5 is to formed out of 6 gents and 4 ladies. In how many ways can this be done when each committee may have at the most two ladies?

Dipak Vitthal Bhawari

Studied at Sinhgad Institute Of Technology, Lonavala5y

At least 2 women means we can choose 2,3,4women

1. 2 women's, 3 men ie 4c2*6c3=6*20=120

2. 3W, 2M ie 4c3*6c2=4*15=60

3. 4W,1M ie 4c4*6c1=1*6=6

120+60+6=186 committee can form

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Venkatesh Krishna Kumar

Mathematics keeps me alive.4y

Related

A committee of 5 is to be selected out of 6 men and 4 women. In how many ways can this be done when at least 2 ladies are included?

Originally Answered: A committee of 5 is to be selected out of 6 men and 4 women .In how many ways can this be done when at least 2 ladies are included?

Given that we need to select at least 2 women and a total of 5, there are 3 cases possible

Case 1: 3 men out of 6, 2 women out of 4 => 6C3 * 4C2 ways = 120

Case 2: 2 men out of 6, 3 women out of 4 => 6C2 * 4C3 ways = 60

Case 3: 1 men out of 6, 4 women out of 4 => 6C1 * 4C4 ways = 6

Total 186 ways possible

Guys, does anyone know the answer?