How many palindromes of length 5 can you form using letters with the following properties: they start with a consonant, and the consonants and vowels alternate; no letter appears more than twice. (Note: assume letters "a", "e", "i", "o", and "u" are the vowels of the English alphabet).

  2100

Step-by-step explanation:

There are 21 consonants that can serve as the first and last letters. There are 5 vowels that can serve as the 2nd and 4th letters. There are 20 remaining consonants that can serve as the 3rd letter. (The same consonant cannot appear in all three places.)

So, the total number of 5-letter palindromes that start with a consonant and alternate with a vowel will be ...

  21×5×20 = 2100 . . . different palindromes

step-by-step explanation:

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)