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# what are the different ways of doing this

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what are the different ways of doing this [#permalink]

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14 Apr 2005, 18:32
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what are the different ways of doing this?
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VP
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14 Apr 2005, 18:45
i have these:

(i) (7C1) (4C2) + (7C2)(4C1) + (7C3)(4C0) = 161
(ii) 11C3 - 4C3 = 161
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14 Apr 2005, 20:28
MA wrote:
i have these:

(i) (7C1) (4C2) + (7C2)(4C1) + (7C3)(4C0) = 161
(ii) 11C3 - 4C3 = 161

Yes 161ways.I followed the 1st method of MA
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14 Apr 2005, 22:16
# of ways to form groups of 3 = 11!/3!8! = 165 ways
# of ways to form groups of 3 with no boys is to pick 3 girls from among the 4 girls and the # of ways to do that is = 4!/3!1! = 4 ways
So # of ways to form groups of 3 with at least 1 boy = 165 -4 = 161 ways
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15 Apr 2005, 04:44
mirhaque wrote:
what are the different ways of doing this?

NOW WHAT WOULD BE THE ANSWER, IF THE GROUP ALWAYS INCLUDE A BOY NAMED JOHN.
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15 Apr 2005, 16:58
mirhaque wrote:
mirhaque wrote:
what are the different ways of doing this?

NOW WHAT WOULD BE THE ANSWER, IF THE GROUP ALWAYS INCLUDE A BOY NAMED JOHN.

Case 1:1B (John) + 2G = C(4,2) = 6 ways
Case 2:2B (John + 1B) + 1G = C(6,1)xC(4,1) = 6x4 = 24 ways
Case 3:3B (John + 2B) + 0G = C(6,2) = 15 ways

Thus total = 45 ways.
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Re: combination   [#permalink] 15 Apr 2005, 16:58
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