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
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 = 13C7 = = 1716.
Hence, 1716.
Workspace:
4 men & 3 women in the team?
Answer: 700
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Explanation :
Number of ways of selecting 4 men out of 7 = 7C4 = 35 ways.
Number of ways of selecting 3 women out of 6 = 6C3 = 20 ways.
∴ Total number of ways of selecting 4 men and 3 women = 35 × 20 = 700 ways.
Hence, 700.
Workspace:
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 1st question)
Hence, 1716.
Workspace:
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 = 7C7 = 1
∴ Total number of ways of selecting one person from each gender = 1716 – 0 – 1 = 1715.
Hence, 1715.
Workspace:
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 = 6C4 × 7C3 = 35 × 15 = 525.
Case 2: 5 women and 2 men are selected.
Number of ways = 6C5 × 7C2 = 6 × 21 = 126.
Case 3: 6 women and 1 man are selected.
Number of ways = 6C6 × 7C1 = 1 × 7 = 7.
∴ Number of ways of selecting women in majority = 525 + 126 + 7 = 658.
Hence, 658.
Workspace:
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 = 12C6 = 924.
Hence, 924.
Workspace:
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 = 12C7 = 792.
Hence, 792.
Workspace:
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 1st question)
Now, if P & E are to be chosen then we need to select 5 more members from the remaining 11 people = 11C5 = 462.
∴ Total number of ways of selecting the team such that P & E both are not chosen together = 1716 - 462 = 1254
Hence, 1254.
Workspace:
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 = 11C5 = 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 = 12C6 = 924.
∴ Total number of ways = 462 + 924 = 1386.
Hence, 1386.
Workspace:
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 = 3C2 = 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 = 4C2 = 6 ways.
∴ Total number of ways = 3 × 6 = 18.
Hence, 18.
Workspace:
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