Need help with a few questions.

Hey guys, I need help with a few questions I have to do for an assignment.

1. Radio and television stations in the USA and Canada have names that are basically four letter words. These words must begin with a K, a W, or a C and may use a blank letter as the last letter. For instance both WHO and CKLW are acceptable names.

a) How many such station names are possible?

b) How many of these are three-letter names (the last letter is a blank)?

c) How many of these names use the same letter throughout the name?

2. A bridge hand consists of 13 cards. How many bridge hands include 5 cards of one suit, 6 cards of a second, and 2 cards of a third?

Have to use combinations, permutations, or factorials for these. I don't even know where to start.

Thanks for the help in advance!

Re: Need help with a few questions.

One problem (at most two) per thread please. Also, show your work or describe your difficulty.

Re: Need help with a few questions.

Quote:

Originally Posted by

**emakarov** One problem (at most two) per thread please. Also, show your work or describe your difficulty.

I've removed one problem.

Please, any help is appreciated.

Re: Need help with a few questions.

How many two-letter words start with W? What about three- and four-letter words?

Re: Need help with a few questions.

Quote:

Originally Posted by

**emakarov** How many two-letter words start with W? What about three- and four-letter words?

26 2 letter words start with W.

So 26P2 for 3 letter words = 650

And 26P3 for 4 letter words = 358000 EDIT: 15600

Is that correct?

Re: Need help with a few questions.

Correct for two letters. For three letters, it is necessary to use 3-permutations when letters cannot be used more than once. There is no such restriction here. So, it's 26^2 for three letters and 26^3 for four letters.

For each starting letter (W, K and C), you need consider the total number of three- and four-letter words. Then multiply the result by 3 to get the total number of possible station names.

Re: Need help with a few questions.

Quote:

Originally Posted by

**emakarov** Correct for two letters. For three letters, it is necessary to use 3-permutations when letters cannot be used more than once. There is no such restriction here. So, it's 26^2 for three letters and 26^3 for four letters.

For each starting letter (W, K and C), you need consider the total number of three- and four-letter words. Then multiply the result by 3 to get the total number of possible station names.

Okay so I got 3(26^2 + 26^3) = 54756 which would be the total # of possible station names.

Then for b) would the answer be 26^2 = 676?

And for c) I don't know what to do.

Re: Need help with a few questions.

For 2), you need to pick 3 of the 4 suits to represent your hand. How many ways can you do this?

e.g: {Club, Spade, Heart} or {Diamond, Heart, Club} (order of suits is NOT important)

The rest isn't too bad once you get this

Re: Need help with a few questions.

Quote:

Originally Posted by

**zerdus** Okay so I got 3(26^2 + 26^3) = 54756 which would be the total # of possible station names.

Yes.

Quote:

Originally Posted by

**zerdus** Then for b) would the answer be 26^2 = 676?

No, this is only the three-letter stations that start with W (or with K, or with C).

Quote:

Originally Posted by

**zerdus** And for c) I don't know what to do.

Hmm, can you write several names that use the same letter throughout?

Re: Need help with a few questions.

Quote:

Originally Posted by

**emakarov** Yes.

No, this is only the three-letter stations that start with W (or with K, or with C).

So 3(26^2) = 2028?

Quote:

Originally Posted by

**emakarov** Hmm, can you write several names that use the same letter throughout?

WWW, KKK, CCC

&

WWWW, KKKK, CCCC

So 6?

Re: Need help with a few questions.

Quote:

Originally Posted by

**Jame** For 2), you need to pick 3 of the 4 suits to represent your hand. How many ways can you do this?

e.g: {Club, Spade, Heart} or {Diamond, Heart, Club} (order of suits is NOT important)

The rest isn't too bad once you get this

So would it be 4C3? (4 suits, how many combinations of 3?)

4C3 is 4, then where do I go from there?

Re: Need help with a few questions.

Quote:

Originally Posted by

**zerdus** So 3(26^2) = 2028?

Yes for three-letter names.

Quote:

Originally Posted by

**zerdus** WWW, KKK, CCC

&

WWWW, KKKK, CCCC

So 6?

Yes.

Re: Need help with a few questions.

Quote:

Originally Posted by

**zerdus** So would it be 4C3? (4 suits, how many combinations of 3?)

4C3 is 4, then where do I go from there?

Yep, it is 4C3.

Now what you have to do is pick 5 of one suit, 6 of another suit, and 2 of the final suit.

Remember, each suit has 13 cards.

Re: Need help with a few questions.

Quote:

Originally Posted by

**Jame** Yep, it is 4C3.

Now what you have to do is pick 5 of one suit, 6 of another suit, and 2 of the final suit.

Remember, each suit has 13 cards.

Hmmm..

So 4C3 * 5? And 4C3 * 6? And 4C3 *7?

Re: Need help with a few questions.

Quote:

Originally Posted by

**emakarov** Yes for three-letter names.

Yes.

Thanks a bunch for the help on that question!