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permutations & combinations

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permutations & combinations

by mkhanna » Mon Jul 13, 2009 8:14 am
A law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which atleast one of the members of the group is a senior partner? (2 groups are considered to be different if atleast one of the group members is different)

Answer choices:
a) 48
b) 100
c) 120
d) 288
e) 600
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by tom4lax » Mon Jul 13, 2009 8:48 am
Three different possibilites

1) 3 Senior Partners + 0 Junior Partners
4!/(3!1!) x 6!/(6!0!) = 4 x 1 = 4

2) 2 Senior Partners + 1 Junior Partner
4!/(2!2!) x 6!/(5!1!) = 6 x 6 = 36

3) 1 Senior Partner + 2 Junior Partners
4!/(3!1!) x 6!/(4!2!) = 4 x 15 = 60

4 + 36 + 60 = 100

IMO answer is B.
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by cata1yst » Mon Jul 13, 2009 11:49 am
tom4lax wrote:Three different possibilites

1) 3 Senior Partners + 0 Junior Partners
4!/(3!1!) x 6!/(6!0!) = 4 x 1 = 4

2) 2 Senior Partners + 1 Junior Partner
4!/(2!2!) x 6!/(5!1!) = 6 x 6 = 36

3) 1 Senior Partner + 2 Junior Partners
4!/(3!1!) x 6!/(4!2!) = 4 x 15 = 60

4 + 36 + 60 = 100

IMO answer is B.

I got the same answer the same way...
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by mike22629 » Mon Jul 13, 2009 12:06 pm
First find total possibilities with all 10 members.

10!/(7!*3!) = 120

Then find all possibilies with NO senior members.

6!/(3!*3!) = 20

120 - 20 = 100
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by rangerguy2000 » Tue Jul 14, 2009 10:19 am
[quote="mike22629"]First find total possibilities with all 10 members.

10!/(7!*3!) = 120

Then find all possibilies with NO senior members.

6!/(3!*3!) = 20

120 - 20 = 100[/quote]


Great way to solve it quickly ... i was doing it the other way too... kudos
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Re: permutations & combinations

by doclkk » Tue Jul 14, 2009 3:02 pm
mkhanna wrote:A law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which atleast one of the members of the group is a senior partner? (2 groups are considered to be different if atleast one of the group members is different)

Answer choices:
a) 48
b) 100
c) 120
d) 288
e) 600
You can quickly rule out 3 and get 50/50 odds.

10C3

10! / 7!3! = 120

That's 120 groups of 3 no restrictions. So Eliminate C/D/E

48 / 100 are left.

In this instance 100 is probably the smarter guess of the two because based on the question and your knowledge of basic combinatorics. you probably wouldn't eliminate 72 groups even though you might have no idea how to do the math of eliminating the groups with restrictions.

Can someone go over the math if the question was at least one junior partner and not senior partner through the Unrestricted minus Restricted method?

Thanks
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