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Find the number of permutations that can be formed from the letters of the word "POPULAR". How many if these permutations,

i)begin and end with "P"?

ii)have two "P" separated?

iii)have the vowels together?

Can anyone explain to me how to get it?

For (i), is it $2P2\cdot 5P5\cdot 2P1$?

For (ii) and (iii), I seriously have no idea ><

2 Answers 2

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(i) Permute 5 inner remaining letters: $5!$

(ii) Subtract from the total cases the number of permutations that have "PP": $\frac{7!}{2!} - 6*5!$

(iii) Group the vowels together and permute each set: $\frac{5*3!4!}{2!}$

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    Hi, thanks for the answers. I'm just a beginner who's learning Stats. So my brain might function...kinda bad.... does 6*5 means 6 multiply 5? ^^;; And I dont really understand (ii) & (iii) >< – 2017-02-14
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    Does that means (i) is 120? >< I'm abit confuse why do we only permute the inner one ;; – 2017-02-14
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    Yes, 6*5 is the same as multiplying 6 with 5. In this case, 6*5! = 6! – 2017-02-14
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(i) This means the two P's are fixed. Hence, we only need to find the permutations of OULAR. There are $5!$ such permutations.

(ii) We count the total amount of permutations. Then we substract the amount of permutations with 2 P's together. This will give us the desired amount:

Thus, the total amount of permutations is $\frac{7!}{2!}$ The amount of permutations with P together looks like this:

PPOULAR

We can consider the PP as one letter as it has to stay together. Call this letter S.

We have to find the permutations of SOULAR: There are $6$! such permutations.

Hence, the answer is $\frac{7!}{2!} - 6!$

(iii) We want the vowels together. Thus we have something that looks like:

POUAPLR

Call the group OUA with the letter S. We can arrange the group S in $3!$ ways, we can arrange PSPLR in $\frac{5}{2}$ ways. Hence, the answer is $\frac{5!*3!}{2!}$

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    Thanks for the explainations ! But, may I know, although the two P are fixed, ain't we suppose to permute it as in 2P2 X 5! X 2P1 ? – 2017-02-14
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    The point is, that if we know that the first letter is a P and the last letter is a P, the only options we have to form different words is to change what happens between those 2 letters. Thus, leaving us with a permutation of 5 letters. – 2017-02-14
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    Okay, noted with thanks! – 2017-02-14