# Need help with a few questions.

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• Feb 24th 2013, 11:23 AM
zerdus
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!
• Feb 24th 2013, 11:51 AM
emakarov
Re: Need help with a few questions.
• Feb 24th 2013, 11:54 AM
zerdus
Re: Need help with a few questions.
Quote:

Originally Posted by emakarov

I've removed one problem.

• Feb 24th 2013, 11:56 AM
emakarov
Re: Need help with a few questions.
• Feb 24th 2013, 11:59 AM
zerdus
Re: Need help with a few questions.
Quote:

Originally Posted by emakarov

So 26P2 for 3 letter words = 650
And 26P3 for 4 letter words = 358000 EDIT: 15600

Is that correct?
• Feb 24th 2013, 12:08 PM
emakarov
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.
• Feb 24th 2013, 12:16 PM
zerdus
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.
• Feb 24th 2013, 12:18 PM
Jame
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
• Feb 24th 2013, 12:20 PM
emakarov
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?
• Feb 24th 2013, 12:26 PM
zerdus
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?
• Feb 24th 2013, 12:29 PM
zerdus
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?
• Feb 24th 2013, 12:29 PM
emakarov
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.
• Feb 24th 2013, 12:32 PM
Jame
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.
• Feb 24th 2013, 12:49 PM
zerdus
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?
• Feb 24th 2013, 12:57 PM
zerdus
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!
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