How many words from $5$ different letters can be formed out of $10$ consonants and $5$ vowels if 'a' is always one of the vowels and the words have at least $2$ consonants?

My approach to this has been like this:

$1$ way of choosing "a".

${10\choose 2}$ ways of choosing the two mandatory consonant.

${14\choose 2}$ ways of choosing the letters for the other two letters of the word.

$5!$ ways of rearranging everything.

So the answer is: $1{10\choose 2}{14\choose 2}5!$

Is this correct?

I think another approach could be calculate all the possible words and subtract the combinations for one vowel and the combinations from $2$ consonants, but I don't get the same result.

  • $\begingroup$ I don't understand the $14$. If you have excluded the $a$ and the two mandatory constants, you should be down to $12$ remaining. Also: You are double counting some combinations. Choosing $B,C$ as the mandatory constants and then choosing $F,G$ as the optional ones is the same as choosing $F,G$ as the mandatory ones and $C,D$ as the optional ones. $\endgroup$
    – lulu
    Commented Sep 26, 2018 at 14:03
  • 2
    $\begingroup$ Best to go by the number of constants. If you have exactly two then there are $\binom {10}2$ ways to choose them and $\binom 42$ ways to choose the extra vowels. If you have exactly three constants then it is $\binom {10}3$ and $\binom 41$ respectively. And so on. $\endgroup$
    – lulu
    Commented Sep 26, 2018 at 14:05
  • $\begingroup$ Why we should consider the ${4\choose 2}$ extra vowels? It is not included in the ${14\choose 2}$ letters for the two additional letters? $\endgroup$
    – Grouper
    Commented Sep 26, 2018 at 15:39
  • 1
    $\begingroup$ If I have a chosen the $a$ and exactly two consonants, then I have to choose exactly two vowels, hence $\binom 42$. I have no idea where your $14$ comes from. $\endgroup$
    – lulu
    Commented Sep 26, 2018 at 17:49
  • $\begingroup$ From all the letters needed to fill the other spaces. $15$ letters in total, minus the "a". $14$. $\endgroup$
    – Grouper
    Commented Sep 26, 2018 at 23:38

2 Answers 2


How many words with five different letters can be formed from ten consonants and five vowels if $a$ is always one of the vowels and the word contains at least two consonants?

Method 1: Consider cases, depending on the number of consonants. Since $a$ must be included, if you have at least two consonants, then the word contains either two, three, or four consonants, with the rest of the letters being vowels. A five-letter word with exactly $k$ of the ten consonants and an $a$ must contain $4 - k$ of the other four vowels. Hence, the number of words with exactly $k$ consonants is $$\binom{1}{1}\binom{10}{k}\binom{4}{4 - k}5! = \binom{10}{k}\binom{4}{4 - k}5!$$ Therefore, the number of admissible words is $$5!\sum_{k = 2}^{4} \binom{10}{k}\binom{4}{4 - k} = 5!\left[\binom{10}{2}\binom{4}{2} + \binom{10}{3}\binom{4}{1} + \binom{10}{4}\binom{4}{0}\right]$$

What did you do wrong?

In your attempt you chose $a$, two consonants, and two additional letters, then arranged them. Since those letters must be different from those you have already selected, you only have $15 - 1 - 2 = 12$ (not $14$) letters left from which to choose. Had you noticed that your approach would have produced the answer $$\binom{1}{1}\binom{10}{2}\binom{12}{2}5!$$ However, that is still wrong. The reason is that by designating two of the consonants as the two consonants that must be included in the word, you count each arrangement with three consonants three times, once for each of the $\binom{3}{2}$ ways you could designate two of those three consonants as the designated consonants, and each arrangement with four consonants six times, once for each of the $\binom{4}{2}$ ways you could designate two of those four consonants as the designated consonants. Notice that $$5!\left[\color{red}{\binom{2}{2}}\binom{10}{2}\binom{4}{2} + \color{red}{\binom{3}{2}}\binom{10}{3}\binom{4}{1} + \color{red}{\binom{4}{2}}\binom{10}{4}\binom{4}{0}\right] = \binom{10}{2}\binom{12}{2}5!$$

Method 2: There are $\binom{15}{5}$ ways to select five of the fifteen available letters. From these, we subtract those that do not contain an $a$, of which there are $\binom{14}{5}$, those that do not contain a consonant, of which there are $\binom{5}{5}$, and those that contain a $a$ and only one consonant, of which there are $\binom{1}{1}\binom{10}{1}\binom{4}{3}$. We then multiply by the $5!$ ways of arranging five distinct letters, which yields $$5!\left[\binom{15}{5} - \binom{14}{5} - \binom{5}{5} - \binom{1}{1}\binom{10}{1}\binom{4}{3}\right]$$

  • 1
    $\begingroup$ Thank you for your insight. I'm having serious difficulties studying this topic on my own. $\endgroup$
    – Grouper
    Commented Sep 27, 2018 at 3:13

Here is an outline toward the solution, which I hope is helpful.

First, you need to be sure to pick the A. You can argue that the probability of picking the A and 4 non-A's in 5 draws without replacement is $1/3.$

Second, given that you have picked the A, @lulu has shown how to find the probability of picking 2 or more consonants.

Then $P(\text{A & 2 Cons})=P(\text{A})P(\text{2 Cons}|\text{A}) = 0.3196803.$

By the simulation below (using R) of a million such choices of five letters, we see that the answer is about 0.319 or 0.320.

ltr = c(1, rep(2,4), rep(3,10))  # 1=A, 2=Other Vowel, 3=Consonant
m = 10^6; a = c = numeric(m)
for (i in 1:m) {
  pick = sample(ltr, 5)
  a[i] = sum(pick==1)
  c[i] = sum(pick==3) }
mean(a==1 & c>=2)
[1] 0.319779

(1/3)*(choose(10,2)*choose(4,2) + choose(10, 3)*choose(4,1) +
[1] 0.3196803

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