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In this problem, we are constructing 12-letter words with four X’s, four Y’s, and four Z’s.

(a)How many such words have no X’s in the first 4 letters?

 

 

Please explain how you got the answers I am curious to know :)

 Mar 17, 2023
edited by Guest  Mar 17, 2023
 #1
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If no X's are allowed in the first 4 letters, then we have to choose all 4 letters from the set {Y, Z} since we need to use 4 of each letter. We have 2 options for each of the first 4 letters, so there are $2^4$ ways to choose these letters. The remaining 8 letters can be any permutation of the 4 X's, 4 Y's, and 4 Z's. There are 12!/(4!4!4!) ways to arrange these 12 letters in total, so the number of 12-letter words with no X's in the first 4 letters is:

$2^4 \times \frac{12!}{4!4!4!} = 5,505,024$.

Therefore, there are 5,505,024 such words with no X's in the first 4 letters.

 Mar 17, 2023
 #3
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Hi! Thanks so much for the quick response!!

 

I don't really undestand how the number of the possibilities for no X's in the first four letters could be bigger than the total possibilities?

Please explain because that will help so much!

 

Thanks so much anyways :)

Guest Mar 18, 2023
 #2
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There are: 12! / [4!4!4!] = 34,650 - total number of 12-digit permutations.

 

Since there are no X's in the first 4 letters, the first 4 letters of the word must consist of only Y's and Z's. There are 8 such letters, so we can choose any 4 of them to be Y's and the remaining 4 to be Z's. The number of ways to do this is:

C(8,4) = 70

 

Once we have selected the first 4 letters, we can arrange the Y's and Z's in any order in the remaining 8 positions. There are C(8,4) ways to choose the positions of the Y's, and the Z's will occupy the remaining positions. Therefore, the number of such 12-letter words is:

 

70 * C(8,4) = 70 * 70 = 4,900

So there are 4,900 such words that have no X's in the first 4 letters.

 Mar 18, 2023

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