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

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Math Expert
Joined: 02 Sep 2009
Posts: 51035
A mathematics department has 5 members. The department must send a tea  [#permalink]

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25 Jul 2018, 20:30
00:00

Difficulty:

15% (low)

Question Stats:

83% (01:05) correct 17% (01:12) wrong based on 81 sessions

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A mathematics department has 5 members. The department must send a team of 2 members to a conference. If there is one pair of members who refuse to attend together, how many possible pairs of department members can be chosen?

A. 14

B. 9

C. 7

D. 5

E. 3

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Joined: 27 Jan 2018
Posts: 42
Re: A mathematics department has 5 members. The department must send a tea  [#permalink]

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25 Jul 2018, 20:34
5c2-2c2=9...i think it is option B??

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2260
A mathematics department has 5 members. The department must send a tea  [#permalink]

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Updated on: 25 Jul 2018, 22:05

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.

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Originally posted by EgmatQuantExpert on 25 Jul 2018, 21:51.
Last edited by EgmatQuantExpert on 25 Jul 2018, 22:05, 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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25 Jul 2018, 21:59
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

Hence, total ways=9

Hence, the correct answer is option B.

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Intern
Joined: 09 Apr 2018
Posts: 31
GPA: 4
Re: A mathematics department has 5 members. The department must send a tea  [#permalink]

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07 Aug 2018, 22:58
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

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

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22 Aug 2018, 22:18
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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22 Aug 2018, 22:57
Total number of pairs = 5C2 = 10
One pair cannot attend together out of 10 = 10 - 1 = 9

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