Combinations ..duh!

[mmukher]

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

[Musiq]

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

The key to most of these problems lies in the language "At Least" or "Exactly" or "No more than".

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.

[mmukher]

Thanks !. I think you should write the test , you sound ready

[Musiq]

Thanks !. I think you should write the test , you sound ready

Thanks mmukher.

I am looking to take the test maybe in September or so. Its more of a deadlines thing.