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Math Expert
Joined: 02 Sep 2009
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A mathematics department has 5 members. The department must send a tea
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25 Jul 2018, 21:30
Question Stats:
80% (01:16) correct 20% (01:19) wrong based on 89 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
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Re: A mathematics department has 5 members. The department must send a tea
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25 Jul 2018, 21:34
5c22c2=9...i think it is option B?? Sent from my SMJ700F using GMAT Club Forum mobile app



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Joined: 04 Jan 2015
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A mathematics department has 5 members. The department must send a tea
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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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Re: A mathematics department has 5 members. The department must send a tea
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25 Jul 2018, 22:59
Method2 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= 101 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
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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: 101=9
Answer B



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






