Combinatorics

[bobdylan]

A company has 3 directors and 5 managers. How many different committees with 5 people can be chosen having at least one director?
a) 25
b)55
c) 500
d)720
e) 4500

[Ashujain]

A company has 3 directors and 5 managers. How many different committees with 5 people can be chosen having at least one director?
a) 25
b)55
c) 500
d)720
e) 4500

No. of ways in which a committee of 5 can be selected from 8 people = 8C5 = 8*7*6/3*2*1 = 56---(1)
No. of ways in which a committee of 5 can be selected with no manager = 5C5 = 1---(2)
Therefore, No. of ways in which a committee of 5 can be selected with at least 1 manager = (1)-(2) = 55

IMO: B

[Brent@GMATPrepNow]

A company has 3 directors and 5 managers. How many different committees with 5 people can be chosen having at least one director?
a) 25
b)55
c) 500
d)720
e) 4500

Questions with "at least" are typically solved faster by first ignoring the restriction and then subtracting the number of outcomes that break the restriction.

For this question, the # of committees with least one director = (the total # of committees possible if we allow any number of directors) - (the # of committees with zero directors)

Total # of committees possible if we allow any number of directors
Three are 8 people and we want to select 5 of them.
This can be accomplished in 8C5 ways (56 ways)

(The # of committees with zero directors)
In other words, we need to select all managers.
There are 5 managers, and we want to select 5 of them.
This can be accomplished in 5C5 ways (1 way)

So, the # of committees with least one director = [spoiler]56 - 1 = 55 = B[/spoiler]

Cheers,
Brent