help with 3 problems

1. If I get what you're asking...

So we need 5 questions for a test, and at least 3 must be multiple choice. We have 20 multiple choice and 10 T/F to choose from.

So we need 3 mc, choosing from 20, so that gives 20C3 (Combination, 20 choose 3). Then, for the other 2 questions, we can pick from the remaining 17mc and 10 T/F, so I think the answer is

20C3 x 27C2.

2. The answer is certainly 6. Any of the 3 ways to chicago can be paired with the plance to cleveland. Likewise, any of them can be paired with the car to cleveland.
 
the first one just so happened to be multiple choice, sorry should of put that. the choices are
51300 99750 115254 15504

so i worked it out like u said
20c3=20!/17!3!=2280
27c2=27!/25!2!=351
351x2280=800280...and that is not a choice...am i doing what u said wrong???????
and ty for helping!
 
Sorry, my bad. I'm not quite sure why my initial approach failed, but here's how you do it:

You want at least 3 mc, which means you want 3, 4, or 5 mc.

How many ways to get (exactly) 3 mc? This means you must have 2 T/F, so it's

20C3 x 10C2

How about 4 mc and 1 T/F?

20C4 x 10C1

and finally, for 5 mc, it's just

20C5

Multiply these quantities and add them up and you should get one of those answers.

I am not quite sure why the first way didn't work, though.
 
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