# How many 10 letter sequences are there using 5 vowels and 5 consonants?

How many 10 letter sequences are there using 5 vowels and 5 consonants? What is the pribability ome of these words has no consecutive pair of consonants?

For the first part I reasoned that each consanta amd vowel could be chosen repeatedly thus: $(21^5)(5^5)$

Now the second part I have been finding tricky:

to rephrase the question I asked: how many different arrangements where a vowel amd consonant alternate such as "vcvcvcvcvc" where v is vowel and c is consonant. Using this idea as the template i initially thought something alomg the lines of $(5)(21)(5)(21)....$ and then multiply by 2 to account for the fact that either my vowel or consonant could be first. But doing this would make my numerator larger than my denominator with a value of : $(21^5)(5^5)(2)$

obviously something wemt wrong in my decomposition.

Suggestions?

• What you have calculated in the first part is the number of ways the first 5 letters can be consonants and the 5 next letters vowels. Commented Dec 6, 2015 at 18:37
• Yep. When reading the problem discern that you are being asked about p(a) OR p(b) which should inform how you construct your expression. Commented Dec 6, 2015 at 18:47

HINT: For the first part of the question, remember that you also need to account for the fact that order matters in the sequence.

Fixing this should also fix your problem with the second part of the question.

• if order matters in this instance what i think you are saying is that since 5 of tge places need to be either a consonant or vowel, would i multiply by $\binom{10}{5}$ for the first part? Commented Dec 6, 2015 at 19:03
• @dc3rd That is correct. Commented Dec 6, 2015 at 20:10

Judging by the comment you placed below Asker's answer, you are now aware that the number of sequences with five vowels and five consonants in which letters may be repeated is $$\binom{10}{5}21^55^5$$ since there are $\binom{10}{5}$ ways to place the five consonants, $21$ choices for each pf the five consonants, and $5$ choices for each of the five vowels.

A sequence of five consonants and five vowels that contains no consecutive consonants and vowels does not necessarily have to alternate consonants and vowels. For instance, BACEIDOFUG is a sequence containing five vowels and five consonants in which no two consonants are consecutive even though the consonants and vowels do not alternate.

To create a sequence of five vowels and five consonants in which no two consonants are consecutive, begin with a sequence of five vowels. We can create $5^5$ such sequences if vowels may be repeated since there are five choices for each vowel. A given sequence of five vowels has six gaps in which we can place the five consonants, four gaps between successive vowels and the two ends of the sequence. We must choose five of these six gaps in which to place the consonants, which can be done in $\binom{6}{5}$ ways. Since consonants may be repeated, we have $21$ choices for each of the five consonants. Thus, the number of sequences of five vowels and five consonants in which no two consonants are consecutive is $$\binom{6}{5}21^55^5$$

To calculate the probability that you seek, divide the second result by the first.

• Thank you for the reasoning behind the second part, that is more important to me than anything else. The way to decompose the question. Commented Dec 6, 2015 at 21:03