Cricket - Combinatorics

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Mozartain: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Tue Nov 11, 2008 8:38 pm; Thanked: 3 times; GMAT Score:740

by Mozartain » Thu Dec 25, 2008 7:14 am

In the last test match against England, the management of Indian cricket team had 16 players to choose from for the playing eleven. These 16 players included 7 bowlers, 7 batsmen, 1 wicketkeeper and 1 allrounder who could both bat and keep wicket. If the team management wanted at least 4 but not more than 5 bowlers and at least 1 but not more than 2 wicketkeepers in the playing eleven, how many such teams could be formed?

*Allrounder = someone who can both bat and bowl, or both bat and keep wicket, or both bowl and keep wicket, or can do all of these.

Source: Beat The GMAT — Problem Solving | Thu Dec 25, 2008 7:14 am

ronniecoleman: Legendary Member; Posts: 546; Joined: Sun Nov 16, 2008 11:00 pm; Location: New Delhi , India; Thanked: 13 times

by ronniecoleman » Thu Dec 25, 2008 8:24 am

WOW~ Cricket.... We Indians love it!!!!

In the last test match against England, the management of Indian cricket team had 16 players to choose from for the playing eleven. These 16 players included 7 bowlers, 7 batsmen, 1 wicketkeeper and 1 allrounder who could both bat and keep wicket. If the team management wanted at least 4 but not more than 5 bowlers and at least 1 but not more than 2 wicketkeepers in the playing eleven, how many such teams could be formed?

First team : 4 bowlers + 1 wicketkeeper + 6 batsman
7C4*2*7C6
Second team: 4 bowlers + 2 wicketkeeper+ 5 batsman
7C4*7C5
Third team: 5 bowlers + 1 wicketkeeper + 5 batsman
7C5*2*7C5
Fourth team 5 bowlers+ 2 wicketkeeper+ 4 batsman
7C5 *7C4

add all !! to get a Dhoni team

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mental: Master | Next Rank: 500 Posts; Posts: 145; Joined: Mon Sep 29, 2008 1:14 am; Thanked: 13 times

by mental » Thu Dec 25, 2008 10:01 pm

SCENARIO 1: 7C4*2C1*8C6
4 bowlers, atleast1-max 2 wicketkeepers, Rest Batsmen

4 bowlers from 7 = 7C4 ways
1 wicketkeeper in 2C1 ways
we have selected 5, required 6 more from 7 batsmen and 1 remaining wicketkeeper = 8C6 ways

SCENARIO 2: 7C5*2C1*8C5
5 bowlers, atleast1-max 2 wicketkeepers, Rest Batsmen

5 bowlers from 7 = 7C5 ways
1 wicketkeeper in 2C1 ways
we have selected 6, required 5 more from 7 batsmen and 1 remaining wicketkeeper = 8C5 ways

Total: 7C4*2C1*8C6 + 7C5*2C1*8C5 = 4312

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