In how many ways can final 11 players be selected from 15 cricket players-

a) there is no restriction

b) one of them must be included

c) one of them who is in bad form must be always be excluded

d) twof them being leg spinners one and only one must be included.

  1. 11 players can be selected out of 15 in 15C11 ways = 15C4 = (15.14.13.12) / (1.2.3.4) = 1365 ways.
  2. Since a particular player must be included, we have to select 10 more out of remaining 14 players. This can be done in 14C10 = 14C4 = (14.13.12.11)/(1.2.3.4) = 1001 ways.
  3. Since a particular player must be always excluded, we have to choose 11 players out of remaining 14. This can be done in14C11 ways = 14C3 = (14.13.12)/(1.2.3) = 364 ways.
  4. One leg spinner can be chosen out of 2 in² C1 = 2 ways. Then we have to select 10 more players out of 13 (because second leg spinner can't be included). This can be done in13C10 ways = 13C3 = (13.12.11)/(1.2.3) = 286 ways. Thus required number of combinations = 2×286 = 572

hope it got it correct

  • 36

1 . 11 players can be selected out of 15 in 15C11 ways = 15C4 = (15.14.13.12) / (1.2.3.4) = 1365 ways.

2. Since a particular player must be included, we have to select 10 more out of remaining 14 players. This can be done in 14C10 = 14C4 = (14.13.12.11)/(1.2.3.4) = 1001 ways.

3. Since a particular player must be always excluded, we have to choose 11 players out of remaining 14. This can be done in14C11 ways = 14C3 = (14.13.12)/(1.2.3) = 364 ways.

4. One leg spinner can be chosen out of 2 in² C1 = 2 ways. Then we have to select 10 more players out of 13 (because second leg spinner can't be included). This can be done in13C10 ways = 13C3 = (13.12.11)/(1.2.3) = 286 ways. Thus required number of combinations = 2×286 = 572

  • 10

Thanx for helping!

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