Permutations & Combinations

I like comedies, animations, and sci fi. A video stores has release wall has 7 sci fi, 2 westerns, 9 suspense, 3 foreign, 12 animation, 4 dramas and 10 comedies. How many way can i choose 3 movies if

b) I must choose one of each type?

c) No more than two of the movies can be sci fi.

b) _{12}C_{1} x _{10}C_{1} x _{7}C_{1} = 12 x 10 x 7 = 840 ways

c) _{7}C_{2} = 21 ... _{42}C_{3} + 21 = 11501 ways

Using all the digits 0,1,2,3,4,5, how many #s less than 4million, and not divisible by either two or five, can be created?

5 x 5 x 4 x 3 x 2 x 1 x 3 = 1800 # can be created

What does B intersect (A union C)

A intersect B' intersect C

(A union C)' intersect B

look like in a Venn Diagram?

How many ways can you make change from $100 bill using only $5 bills and $20 bills?

=2^5 = 32

These are my attempt at these questions, which I don not have the answers to. I would like someone to check if I have done something wrong or if I was on the right track, thank you!

Re: Permutations & Combinations

For the last part use the definition of ^{n}P_{r}=$\displaystyle \frac{n!}{(n-r)!}$

Now it's just simplification of LHS.

Re: Permutations & Combinations

Quote:

Originally Posted by

**MathIsNotFun** I like comedies, animations, and sci fi. A video stores has release wall has 7 sci fi, 2 westerns, 9 suspense, 3 foreign, 12 animation, 4 dramas and 10 comedies. **How many way can i choose 3 movies if**

a) There are no restrictions?

b) I must choose one of each type?

c) No more than two of the movies can be sci fi.

a) Total = 7+2+9+3+12+4+10 = 47 ... _{47}C_{3} = 16215 ways

b) _{12}C_{1} x _{10}C_{1} x _{7}C_{1} = 12 x 10 x 7 = 840 ways

c) _{7}C_{2} = 21 ... _{42}C_{3} + 21 = 11501 ways

I have some concerns about this question.

I agree with you on part a).

However, part b) makes no sense. If you choose only three there is no way to choose one of each.

Please review the actual statement and edit.

Quote:

Originally Posted by

**MathIsNotFun** How many 'words' are possible using all the letters TRANSUBSTANTIATION

18!/1!2!3!2!3! = 18!/3456 = 1.85 x 10^12 words

Here you have miss-counted the number of T's. You need to start over.

Quote:

Originally Posted by

**MathIsNotFun** Using all the digits 0,1,2,3,4,5, how many #s less than 4million, and not divisible by either two or five, can be created?

5 x 5 x 4 x 3 x 2 x 1 x 3 = 1800 # can be created

These are **seven digit** numbers. There are three possibilities for the left most digit and two choices for the unit digit.

How many choices for the other five digits?

Re: Permutations & Combinations

The first question is worded exactly as it is, so Im not sure what the teacher meant :S.

Oh I think I made a typo the 1! supposed to be a 4!.

Hmmm I'll try and correct that one.

New question added :S!