# Counting different way of arranging letters from two groups, without repetitions

We have the vowels a, e, i, o, u, y in one group and the remaining 20 consonants in the other group. The first part of the problem is to count how many words we can form using three letters from each group without repetitions. To solve this, I counted this way: $$20*19*18*6*5*4=820800$$

The following problems are variations of this:

• How many of these words contain both the letters c and i?
• How many of these words contain both the letters c and d?

The issue I'm having here is how to count these possibilities, considering these letters could be located anywhere in the word. Sometimes they are both in the same subgroup, sometimes in different groups. I'm not looking necessarily for the solution to this problem, but rather to what I need to understand in order to solve it.

EDIT: I was wrong in my original calculation and understanding of the problem. Instead of trying to put the letters into order right away, it works better by counting the number of unique sets we can create, then multiply that by the number of ways we can arrange a set of that length.

For example, we start by counting the number of unique sets that we can create using 3 letters out of 20. The order doesn't of the letters within the sets doesn't matter for now. Then, we do the same thing for the vowels, using 3 letters out of 6. I get 1140 unique sets of consonants and 20 unique sets of vowels. Multiplying those two together gives us 22800 unique sets, each containing 3 vowels and 3 consonants. From this point, we just have to count all the ways we can arrange those sets, which is 6! times.

For the following problems, I just had to do the previous process but with more or less vowels and consonants.

How many words of length six can be formed using three of the six vowels a, e, i, o, u, y and three of the twenty consonants if no letter is repeated?

The order of the letters matters. Choose three of the six positions for the vowels. We have six choices for the leftmost vowel, five choices for the middle vowel, and four choices for the rightmost vowel. We fill the remaining positions with consonants. We have twenty choices for the leftmost consonant, nineteen choices for the middle consonant, and eighteen choices for the rightmost consonant. Hence, the number of such words is $$\binom{6}{3} \cdot 6 \cdot 5 \cdot 4 \cdot 20 \cdot 19 \cdot 18$$

How many of these words contain both the letters c and i?

Choose one of the six positions for the c. Choose one of the five remaining positions for the i. Choose two of the remaining four positions for the remaining two vowels. Choose the vowels from the five that remain available. Choose two consonants from the nineteen consonants that remain available to fill the remaining open positions.

How many of these words contain both the letters c and d?

Choose one of the six positions for the c. Choose one of the five remaining positions for the d. Choose the location for the remaining consonant. Choose which of the eighteen remaining consonants will fill that position. Choose three of the six vowels to fill the remaining positions.

• I ended up understanding the problem better by putting the letters in order at the very end. I updated my question to explain my reasoning, but in short, I start by counting the number of different set of letters I can create, then I multiply that number by the number of ways I can arrange them differently. – Samuel Lapointe Mar 11 '17 at 23:03