Out of 16 players in a cricket team, 4 are bowlers and 2 are wicket keepers. A team of 11 players with at least 3 bowlers and 1 wicket keeper is to be formed. Find the number of ways the team can be selected.
My solution:
Choosing 3 bowlers out of 4: $\binom43$
Choosing 1 wicket keeper out of 2: $\binom21$
Choosing the remaining 4 players out of 12: $\binom{12}{7}$
Hence, total permutations=$\binom43\binom21\binom{12}{7}=6336$
Given solution:
Number of bowlers=4
Number of wicket keepers=2
Total required selection=$\binom43\binom21\binom{10}{7} +\binom44\binom21\binom{10}{6}+\binom43\binom22\binom{10}{6}+\binom44\binom22\binom{10}{5}=2472$
I feel my solution is just a simpler version of what the book gives. But why is then my answer coming different and where have I gone wrong in my approach?
Best Answer
Your solution counts some solutions more than once, namely if you have four bowlers or two wicket keeps you count them four times and twice respectively. The book however counts each in a different place: Each term on the LHS is one of four possibilities, depending on how many bowlers/wicket keepers there are.