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