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# Prob by BB

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Manager
Joined: 12 Mar 2003
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Prob by BB [#permalink]

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03 Aug 2003, 14:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

BB, these are your questions. Do you care to explain the answers ?

1. How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used? (Ans. 1260)

2. There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)
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Prob by BB: Solution [#permalink]

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03 Aug 2003, 15:26
1) L - denotes factorial.

L9/(L3 * L4 * L2)

2) The answer coming out for me is 60.

We can have either one man or one woman in the group.
Case 1: One man and two women.
Ways to select a man = 5
Ways to select the first woman = 4
Ways to select the second woman = 3
Total ways = (5*4*3)/2 = 30 [Dividing by 2 bcos the woman can interchange]
Case 2: One woman and two men.
Same calculation : Result = 30
Total Combinations = 30 + 30 = 60
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04 Aug 2003, 00:00
(2) 10 persons, we select 3, 10C3=120 is the number of total groups.
Wrong group is one containing H and W. The matrix is HW_. The last place can be filled with any person out of 8 people left. However, there are 5 pairs HW. So, the total number of wrong groups is 5*8=40

120-40=80
Founder
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Re: Prob by BB [#permalink]

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05 Aug 2003, 18:53
tzolkin wrote:
BB, these are your questions. Do you care to explain the answers ?

1. How many different signals can be transmitted by hoisting 3 red, 4 yellow and 2 blue flags on a pole, assuming that in transmitting a signal all nine flags are to be used? (Ans. 1260)

2. There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)

I guess I am too late; too old.
Re: Prob by BB   [#permalink] 05 Aug 2003, 18:53
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# Prob by BB

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