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# In how many ways can a team of 11 cricketers be chosen from

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Director
Joined: 24 Oct 2005
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In how many ways can a team of 11 cricketers be chosen from [#permalink]

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25 Nov 2005, 05:08
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In how many ways can a team of 11 cricketers be chosen from 6 bowlers, 4 wicket keepers and 11 batsmen to give a majority of batsmen if atleast 4 bowlers are to be included and there is 1 wicket keeper.

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Director
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25 Nov 2005, 11:29
the only possible option is 4 blrs, 1wk and 6 batsmen if i understand the q correctly. This can be done in 6C4x4C1x11C6 =15x4x462=27720 ways

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SVP
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Re: Permutation/ Combination - PS [#permalink]

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25 Nov 2005, 12:21
remgeo wrote:
In how many ways can a team of 11 cricketers be chosen from 6 bowlers, 4 wicket keepers and 11 batsmen to give a majority of batsmen if atleast 4 bowlers are to be included and there is 1 wicket keeper.

if you want to include majority batsman and atleast 4 bowlers then

it can be 4bolwers, 6 batters 1wk
= 6C4*11C5*4C1 =15*462*4 = 27720
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hey ya......

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Director
Joined: 24 Oct 2005
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26 Nov 2005, 04:00
Yes ans = 27720.

1 wicket keeper can be selected in C(4,1) ways
4 bowlres chosen = C(6,4)
Remaining 6 batsmen = C(11,6)

Total possibilities = C(4,1) * C(6,4) * C(11,6) = 27720.

The batsmen has to be majority. So the split cannot be 1WC, 5Bowlers, 5Batsmen.
It can only be 1WC, 4bowlers and 6batsmen.

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26 Nov 2005, 04:00
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# In how many ways can a team of 11 cricketers be chosen from

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