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A mathematics department has 5 members. The department must send a tea

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A mathematics department has 5 members. The department must send a tea  [#permalink]

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New post 25 Jul 2018, 21:30
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

80% (01:16) correct 20% (01:19) wrong based on 89 sessions

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Re: A mathematics department has 5 members. The department must send a tea  [#permalink]

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New post 25 Jul 2018, 21:34
5c2-2c2=9...i think it is option B??

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A mathematics department has 5 members. The department must send a tea  [#permalink]

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New post Updated on: 25 Jul 2018, 23:05

Solution



Given:

    • A mathematics department has 5 members.
    • The department must send a team of 2 members to a conference.
    • One pair of members refuse to attend together.

To find:

    • How many possible pairs of department members can be chosen?

Approach and Working:

Let department members are: A, B, C, D, and E and A and B refuse to attend the conference together.

Hence, Total cases:

    • Select A as select 1 member other than B
      o So, we have to select 1 more member out of 3 members.
      o Number of ways= \(^3c_1\)=3

    • Select B as select 1 member other than A
      o So, we have to select 1 more member out of 3 members.
      o Number of ways= \(^3c_1\)=3

    • Do not select either of A or B
      o So, we have to select 2 more members out of 3 members.
      o Number of ways= \(^3c_2\)=3

Hence, total ways=9

Hence, the correct answer is option B.

Answer: B
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Originally posted by EgmatQuantExpert on 25 Jul 2018, 22:51.
Last edited by EgmatQuantExpert on 25 Jul 2018, 23:05, edited 1 time in total.
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Re: A mathematics department has 5 members. The department must send a tea  [#permalink]

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New post 25 Jul 2018, 22:59
1

Method-2



Let department members are: A, B, C, D, and E and A and B refuse to attend the conference together.

Hence, Total cases in which we A and B can be there in a pair:
    • Both A and B are in the pair OR
    • A is in the pair and B is not in the pair
    • B is in the pair and A is not in the pair
    • Neither of A or B is in the pair.

But, out of all the cases we are not selecting 1 case:- When both A and B are in the pair.

Thus, if we remove the case when both A and B are in the pair from the total pairs, we will get the answer.

    • Total Pairs: \(^5c_2\)
    • Total Pairs when both A and B are in the pair= 1

Thus, required answer= 10-1
Hence, total ways=9

Hence, the correct answer is option B.

Answer: B
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Re: A mathematics department has 5 members. The department must send a tea  [#permalink]

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New post 07 Aug 2018, 23:58
using the formula for k combinations of n, with n=5, k=2:

\(5!/(2!*3!)=10\)

We know that 1 combination out of the 10 possible ones is not allowed, so: 10-1=9

Answer B
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Re: A mathematics department has 5 members. The department must send a tea  [#permalink]

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New post 22 Aug 2018, 23:18
1C1*3C1 + 1C1*3C1+3C2=9

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Re: A mathematics department has 5 members. The department must send a tea  [#permalink]

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New post 22 Aug 2018, 23:57
Total number of pairs = 5C2 = 10
One pair cannot attend together out of 10 = 10 - 1 = 9

Answer: B

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Re: A mathematics department has 5 members. The department must send a tea   [#permalink] 22 Aug 2018, 23:57
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