klueless7825 wrote:
I'm trying to solve the problem using reverse combination approach.Please,Can anyone help.
Need to choose 3 people from 16 so 16C3.
No. of ways to choose 3 people from the same team 4C3.since there are four teams (4C3)^4
16C3 - (4C3)^4
=560-256
=304
First of all, this approach is complicated, convoluted and error prone. I'd go by Bunuel's explanation. Different permutation and combination problems are done using different methods. You just need to pick the most apt one for a specific type.
Anyway,
No. of ways to choose 3 people from the same team 4C3.since there are four teams (4C3)^4
This is not entirely correct.
Number of ways to choose 3 people from the same team: 4C3. Correct
There are four teams: (4C3)^4. Not correct.
Remember, here you need to add the numbers not multiply because you choose from 1 OR choose from 2 OR choose from 3 OR choose from 4.
Secondly, you have missed out another case in which just 1 candidate is picked from 1 team AND 2 candidates are picked from another team. Because, that will also be the NOT one candidate from each team.
So, actual answer could be arrived by this:
\(Total=C^{16}_{3}=560\)
Picking all 3 candidates from just one team:
\(C^4_1*C^4_3=16\)
OR(i.e. +)
Picking 2 candidates from one team AND picking 1 candidate from one of the remaining teams
\(C^4_1*C^4_2*C^3_1*C^4_1=288\)
Total number of ways in which 1 candidate is NOT picked from each team
\(16+288=304\)
Now, to find the left-overs just subtract from total:
\(560-304=256=4^4\)
Ans: "B"