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 :)
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.
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.
