A club as 30 members including 3 lawyers,4 teachers and 5 docters.In how many ways can a committee of 8 be formed to contain 1 teacher,2 lawyers,and 2

[Zai47olali]

A club as 30 members including 3 lawyers,4 teachers and 5 docters.In how many ways can a committee of 8 be formed to contain 1 teacher,2 lawyers,and 2 docters?

 

[Dr.Peterson]

What help do you need? Be sure to read this.

Assuming the question requires exactly 1 teacher, 2 lawyers, and 2 doctors (not at least ...), you need to count the number of ways to choose 1 of the 4 teachers, 2 of the 3 lawyers, 2 of the 5 doctors, and 3 of the 18 others. Do you know how to do that (combinations)?

 

[pka]

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[Steven G]

What help do you need? Be sure to read this.

Assuming the question requires exactly 1 teacher, 2 lawyers, and 2 doctors (not at least ...), you need to count the number of ways to choose 1 of the 4 teachers, 2 of the 3 lawyers, 2 of the 5 doctors, and 3 of the 18 others. Do you know how to do that (combinations)?

Dr P, if it were at least would the solution be the number of ways to choose 1 of the 4 teachers, 2 of the 3 lawyers, 2 of the 5 doctors, and 3 of the remaining 25? I'm thinking that making looking at all of the possible cases would be necessary to solve this?

 

[Dr.Peterson]

Beyond mentioning the possibility of "at least" (and shuddering), I gave zero thought to it, because I think it would be a horrendous inclusion-exclusion process, or else just a huge summation. But real combinatorists (is that the word?) might have a better way ... or they might just use a computer.