# CRE 5 - Forming a committee | Modern Math - Permutation & Combination

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**Answer the next 10 questions based on the information given below: **

A team of seven members is to be formed from among 7 men – A, B, C, D, E, F & G and 6 women – P, Q, R, S, T & U. In how many ways can the committee be formed such that there is/are

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

No restrictions on team selection?

Answer: 1716

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**Explanation** :

Total members available for selection = 7 + 6 = 13.

∴ 7 people are to be selected from 13, hence number of ways = ^{13}C_{7} = $\frac{13\times 12\times 11\times 10\times 9\times 8\times 7}{7\times 6\times 5\times 4\times 3\times 2\times 1}$ = 1716.

Hence, 1716.

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**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

4 men & 3 women in the team?

Answer: 700

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**Explanation** :

Number of ways of selecting 4 men out of 7 = ^{7}C_{4} = 35 ways.

Number of ways of selecting 3 women out of 6 = ^{6}C_{3} = 20 ways.

∴ Total number of ways of selecting 4 men and 3 women = 35 × 20 = 700 ways.

Hence, 700.

Workspace:

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

At least 1 man in the team?

Answer: 1716

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**Explanation** :

Since there are only 6 women but 7 members have to selected, hence at least 1 male member will always be selected in the team.

∴ Total number of ways of selecting 7 members = 1716 (From 1^{st} question)

Hence, 1716.

Workspace:

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

At least one person from each gender?

Answer: 1715

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**Explanation** :

Total number of ways of selecting one person from each gender = Total ways – Ways of selecting only women – Ways of selecting only men.

Total number of ways of selecting only women = 0

Total number of ways of selecting only men = ^{7}C_{7} = 1

∴ Total number of ways of selecting one person from each gender = 1716 – 0 – 1 = 1715.

Hence, 1715.

Workspace:

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

Women are in majority?

Answer: 658

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**Explanation** :

For women to be in majority, number of women to be selected is either 4 or 5 or 6.

**Case 1:** 4 women and 3 men are selected.

Number of ways = ^{6}C_{4} × ^{7}C_{3} = 35 × 15 = 525.

**Case 2:** 5 women and 2 men are selected.

Number of ways = ^{6}C_{5} × ^{7}C_{2} = 6 × 21 = 126.

**Case 3:** 6 women and 1 man are selected.

Number of ways = ^{6}C_{6} × ^{7}C_{1} = 1 × 7 = 7.

∴ Number of ways of selecting women in majority = 525 + 126 + 7 = 658.

Hence, 658.

Workspace:

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

C is always present in any chosen committee?

Answer: 924

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**Explanation** :

If C is always chosen in the team, there are 6 more members to be selected from remaining 12 people.

∴ Number of ways = ^{12}C_{6} = 924.

Hence, 924.

Workspace:

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

F is not chosen for the committee?

Answer: 132

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**Explanation** :

If F is not to be selected, we have to select 7 members from remaining 12 people.

∴ Number of ways = ^{12}C_{7} = 792.

Hence, 792.

Workspace:

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

P & E are not simultaneously chosen for the committee?

Answer: 1254

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**Explanation** :

Total number of ways of selecting the team such that P & E both are not chosen together = Total ways – Ways of choosing both P & E.

Total ways = 1716 (From 1^{st} question)

Now, if P & E are to be chosen then we need to select 5 more members from the remaining 11 people = ^{11}C_{5} = 462.

∴ Total number of ways of selecting the team such that P & E both are not chosen together = 1716 - 462 = 1254

Hence, 1254.

Workspace:

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

S is chosen, only if B is chosen for the committee?

Answer: 1386

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**Explanation** :

**Case 1:** S is chosen. Since S is chosen B also has to be selected. Now we need to select 5 more people from the remaining 11.

Number of ways = ^{11}C_{5} = 462.

**Case 2:** S is not chosen. Now there is no restriction for B. Now we need to select 6 more people from the remaining 12.

Number of ways = ^{12}C_{6} = 924.

∴ Total number of ways = 462 + 924 = 1386.

Hence, 1386.

Workspace:

**CRE 5 - Forming a committee | Modern Math - Permutation & Combination**

2 women & 5 men such that B, C, & G must be chosen while none among Q, T & U must be chosen?

Answer: 18

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**Explanation** :

Out of 6 women i.e., P, Q, R, S, T & U, 2 are to be selected such that Q, T and U are not to be selected.

∴ 2 women are to be selected from P, R and S = ^{3}C_{2} = 3 ways.

Out of 7 men i.e., A, B, C, D, E, F & G, 5 are to be selected such that B, C and G must be selected.

∴ 2 more men are to be selected from A, D, E and F = ^{4}C_{2} = 6 ways.

∴ Total number of ways = 3 × 6 = 18.

Hence, 18.

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