SOLUTION: How many different arrangements can be made of all the letters of the word "ACCOUNTANTS"? In how many of them the vowels stand together?

Question 1190938: How many different arrangements can be made of all the letters of the word
"ACCOUNTANTS"? In how many of them the vowels stand together?

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part 1) How many different arrangements can be made of all the letters of the word
"ACCOUNTANTS"?

The word ACCOUNTANTS has 11 letters.

If we could distinguish the repeated letters from one another, then we'd have 11! = 39916800 different permutations of the letters. The exclamation mark means factorial.
We start at 11 and count down to 1 multiplying along the way
11*10*9*8*7*6*5*4*3*2*1 = 39916800
Optionally, you could use the nPr formula with n = 11 and r = 11 to get the same result.

Whichever path you take, we would have 39916800 different permutations if we could tell the repeated letters apart.

However, we cannot tell the 'A's apart for instance.
Same goes for the N's, the T's and the C'
We have 4 instances where a letter is repeated.
Each instance means we double counted.
Overall, we overcounted by a factor of 2*2*2*2 = 16

The value 39916800 is overcounted by a factor of 16, so we must divide by 16 to fix the error.
In reality, there are 39916800/16 = 2,494,800 different permutations.

Answer: 2,494,800
This is a little bit under 2.5 million

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Part 2) In how many of them the vowels stand together?

Let X be a stand-in for the vowels that will be grouped together as one string.

We'll go from the word ACCOUNTANTS to the "word" XCCNTNTS.
This string of letters results when you delete all the vowels (A O U and A), then stick X up front.

The idea is that arranging the letters of that strange "word" will help us toward the goal of finding the number of permutations where the vowels are lumped together.
This is because wherever you see an X, replace it with some permutation of A,A,O,U.
Example: the permutation CXCNTNTS is the same as CAAOUCNTNTS where X = AAOU in this case.
Another Example: the permutation CXCNTNTS is the same as COUAACNTNTS where X = OUAA in this case.

There are 8 letters in XCCNTNTS
There would be 8! = 40320 permutations if we could tell the repeated letters apart.
However, we cannot tell them apart. So we divide by 2*2*2 = 8 to fix the error (we don't have repeated 'A's this time).
So there are 40320/8 = 5040 permutations of XCCNTNTS

Within any permutation, there are 4!/2 = 24/2 = 12 ways to arrange the four vowels A,A,O,U. I divided by 2 to deal with the repeated "A" vowel.

So we have 12*5040 = 60480 different permutations where we are guaranteed to have the vowels all clumped together in some fashion.

Answer: 60480

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