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(20060)   (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(52804)  (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(13200)  (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
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