Cant seem to get this really easy one.
1> Group of 10 members. 4 French and the rest are Spanish/German. How many ways to choose a 3 member committee, with atleast one French..
My approach (which is wrong Im guessing since it doesnt come up with the OA)
3 spots available, 1 reserved for the French others open to all.
So 4C1 * 9C2
Aother approach (which gives the right answer)
Total combinations are 10C3
Number of ways of selecting NO french = 6C3
Atleast one French = 10C3 - 6C3.
Whats wrong with approach 1 ?
Thanks
Combinations ..duh!
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The key to most of these problems lies in the language "At Least" or "Exactly" or "No more than".mmukher wrote:
1> Group of 10 members. 4 French and the rest are Spanish/German. How many ways to choose a 3 member committee, with atleast one French..
3 spots available, 1 reserved for the French others open to all.
So 4C1 * 9C2
Your assumption was that Atleast 1 implies 1 spot reserved for the French person.
This doesnt account for the 2 other situations of 2 French and 3 French peeps in the committe. Both of these come under the Atleast 1 spectrum.
If you wanted to use this approach, the following would give you the answer, but its laborious:
Let F stand for a Frenchperson (theres 4 of them to pick from), and Others stand for the rest (theres 6 of them to pick from)
Therefore,
All the ways of picking 1 F and 2 Others + All the ways of picking 2 F and 1 Other + All the ways of picking 3 F and Zero others
( AND IS MULTIPLICATION)
This transalates to ( 4C1 ) ( 6C2) +( 4C2 ) ( 6C1) + (4C3) (6C0)
This works out to a 100 ways, which is the same exact answer as the other approach is.
But, the other approach is definitely test-smart and recommended.
For love, not money.
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- Senior | Next Rank: 100 Posts
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Thanks mmukher.mmukher wrote:Thanks !. I think you should write the test , you sound ready
I am looking to take the test maybe in September or so. Its more of a deadlines thing.
For love, not money.