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!

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

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Originally Posted by emakarov

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

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I've removed one problem.

Please, any help is appreciated.

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

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Originally Posted by emakarov

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

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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?

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.

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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.

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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.

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

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Originally Posted by zerdus

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

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Yes.

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Originally Posted by zerdus

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

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

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Originally Posted by zerdus

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

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

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Originally Posted by emakarov

Yes.

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

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So 3(26^2) = 2028?

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Originally Posted by emakarov

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

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WWW, KKK, CCC
&
WWWW, KKKK, CCCC

So 6?

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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

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

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

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Originally Posted by zerdus

So 3(26^2) = 2028?

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Yes for three-letter names.

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Originally Posted by zerdus

WWW, KKK, CCC
&
WWWW, KKKK, CCCC

So 6?

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Yes.

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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?

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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.

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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.

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Hmmm..

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

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Originally Posted by emakarov

Yes for three-letter names.

Yes.

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Thanks a bunch for the help on that question!