The first letter is a vowel and the last letter is another vowel without repetitions?

Algebra ->  Permutations -> SOLUTION: How many different five-letter strings can be formed from the letters A, B, C, D, E (repeats allowed) if the first letter must be a vowel and the last letter must be a consonant?       Log On

Click here to see ALL problems on Permutations Question 1135152: How many different five-letter strings can be formed from the letters A, B, C, D, E (repeats allowed) if the first letter must be a vowel and the last letter must be a consonant? Found 3 solutions by Edwin McCravy, ikleyn, greenestamps: Answer by Edwin McCravy(19236)     (Show Source): You can put this solution on YOUR website! A sample string would be ECADB Choose the MOST RESTRICTIVE things to choose first. Among A, B, C, D, E there are 2 vowels and 3 consonants. We can choose the vowel for the first letter in 2 ways, (A or E) That leaves 1 vowel and 2 consonants. We can choose the consonant for the last letter in 3 ways, (B, C or D) That leaves 3 letters to choose. We can choose the second letter any of 3 ways. That leaves 2 letters to choose. We can choose the third letter either of 2 ways. That leaves only 1 letter to choose. We must choose the fourth letter as that 1 remaining way. Answer: multiply them: 2∙3∙3∙2∙1 = 36 ways. FYI, here are all 36 ways, computer generated: 1. ABCED 2. ABDEC 3. ABECD 4. ABEDC 5. ACBED 6. ACDEB 7. ACEBD 8. ACEDB 9. ADBEC 10. ADCEB 11. ADEBC 12. ADECB 13. AEBCD 14. AEBDC 15. AECBD 16. AECDB 17. AEDBC 18. AEDCB 19. EABCD 20. EABDC 21. EACBD 22. EACDB 23. EADBC 24. EADCB 25. EBACD 26. EBADC 27. EBCAD 28. EBDAC 29. ECABD 30. ECADB 31. ECBAD 32. ECDAB 33. EDABC 34. EDACB 35. EDBAC 36. EDCAB Edwin Answer by ikleyn(46337)     (Show Source): You can put this solution on YOUR website! . The way on how tutor @Edwin McCravy solves the problem is not consistent with the given condition. The solution by Edwin assumes  " no repeats allowed",  in opposing with the condition:  "repeats allowed". Answer by greenestamps(11235)     (Show Source): You can put this solution on YOUR website! 1st letter: must be a vowel -- 2 choices 2nd letter: no restrictions -- 5 choices 3rd letter: no restrictions -- 5 choices 4th letter: no restrictions -- 5 choices 5th letter: most be a consonant -- 3 choices By the fundamental counting principle, the total number of 5-letter strings is the product of the numbers of choices you have for each letter: 2*5*5*5*3 = 750

we have 8 letters in total (_ _ _ _ _ _ _ _ _)

4 vowels - A,A,A,I

4 consonants - M,L,Y,S

First letter must be consonant so we can pick any letter from M,L,Y,S so we have 4 options - (4 _ _ _ _ _ _ _)

Last letter must be a vowel we can pick A or I so only 2 choices

• choosing A
• choosing I

Choosing A

we are left with 6 letters 3 consonants and 3 vowels (A,A,I)

so number of possibilities are (6p6)/2!

Choosing I

we are left with 6 letters 3 consonants and 3 vowels (A,A,A)

so number of possibilities are ($\frac{^6p_6}{3!}$)

so total is 4*(6p6/2! + 6p6/3!) = 1920

Your first question answer should be 6720

Asked 7 years, 1 month ago

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find the number of different 8 letter arrangements that can be made from the letters of the word DAUGHTER (8 letters) so that --All vowels occur together. I thought of a solution which is as follows-- if all the vowels are occurring together then wherever these three are they must be together so let us say they start occurring right from start or even from the second or third letter ((no special case or reason is there )) then no of choices for first letter=3 ////@@3 vowels are there. second letter=2 //@@ 2 vowels are remaining . third letter =1 //@@1 vowel is remaining which we must use. fourth letter =5 choices(D,G,H,T,R) fifth =4 choices • • • • so on till the last letter. so number of different possibilities by fundamental principle of counting= 3×2×1×5×4×3×2×1=5!×3!. But the answer is 6!×3!,which case I am missing in my solution because of which I am getting wrong solution(5!×3!)

asked Oct 6, 2015 at 18:19

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I think your basic method sounds ok, but the details get a little hazy.

Suppose we take $AUE$ as a block. Then you are looking at $6$ characters...namely, $\{D,G,H,T,R,AUE\}$. There are $6!$ ways to permute this collection hence there are $6!$ ways to arrange the letters so that the group $AUE$ occurs in that precise order. Now, of course there are also $3!=6$ ways to permute the vowels and, given any permutation, there are $6!$ ways to arrange all the letters so that the vowels occur in that precise order. Hence $6!^*3!$ ways altogether.

answered Oct 6, 2015 at 18:23

lulululu

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