# How many $3$ letter “words” consisting of at least $1$ vowel and $1$ consonant can be made from the letters of EQUATION?

The word EQUATION contains all five vowels. How many $3$ letter "words" consisting of at least $1$ vowel and $1$ consonant can be made from the letters of EQUATION?

Hi, would anyone be able to check the answer for this? The answers say 540, but I keep getting 270.

My means of working out were taking two cases: one with two vowels and one consonant, and one with one vowel and two consonants.

Thanks

• Whoa. I got 270 too. – BCLC Aug 6 '14 at 3:22
• $8^3=512\lt540$, so even if you ignore the constraints and allow yourself to repeat letters, you can't get the ostensible answer. – Barry Cipra Aug 6 '14 at 3:24
• Hmmm... but Night's answer seems correct. – mathsguy Aug 6 '14 at 3:25

Here is another way - take all three letter words and deduct those containing just vowels or just consonants. This comes to $$8\cdot 7 \cdot 6-5\cdot 4 \cdot 3 - 3\cdot 2 \cdot 1=336-60-6=270$$

If letters are allowed to be repeated the number is $$8^3-5^3-3^3=512-125-27=360$$

Another way of counting is to count the number of possibilities for each pattern of vowels and consonants.

$VVC: 5\times 4 \times 3=60$

$VCV: 5\times 3 \times 4=60$

$VCC: 5\times 3 \times 2=30$

$CCV: 3\times 2\times 5=30$

$CVC: 3\times 5\times 2=30$

$CVV: 3\times 2\times 5=60$

This gives $270$

I've done this longhand for clarity

• Im confused since the answer is 540, not 270 (we're assuming letters aren't allowed to be repeated) – mathsguy Aug 6 '14 at 3:46
• @mathsguy I put up these alternative calculations so that people could have a look and pick holes (if they existed). It seems to me that the advertised "answer" is wrong. – Mark Bennet Aug 6 '14 at 3:49
• The advertised answer is greater than any set of three-letter words you can make from eight letters, which is a dead giveaway that it's wrong. (I'm not the first person to notice that, BTW.) – David K Aug 6 '14 at 3:50
• Now that we've concluded that, is there a reason why it's wrong. It seems right. We choose the 3 letters and group them. Do we ever need to divide by 2 at any point? – mathsguy Aug 6 '14 at 3:53
• Thanks, this makes perfect sense now. – mathsguy Aug 6 '14 at 3:55

Think how many different permutations of three letters we can select from a set of eight letters without replacement. It is $8 \cdot 7 \cdot 6 = 336$. (Right away this tells us the answer is wrong if you select without replacement, because the desired set is a subset of this set of words. But the answer $540$ is just as wrong if you select with replacement, via a similar argument.)

Now how many can you make containing only vowels, if you have five vowels available? That's $5 \cdot 4 \cdot 3 = 60$.

Now how many can be made with the three consonants you have? That's $3 \cdot 2 \cdot 1 = 6$.

But all the three letter words either have one vowel and two consonants, two vowels and one consonant, all vowels, or all consonants. Hence the set you're looking for consists of all words except the ones that are all vowels or all consonants, and the number of words in the set is

$$336 - 60 - 6 = 270.$$

• Thanks for the extra answer. It's important that this gets clarified :D. All these answers seem correct :/ – mathsguy Aug 6 '14 at 3:50
• I started the answer before Mark Bennet posted, but he worked much faster than I did. He gave all the necessary details more succinctly, too. I say give him the check mark. – David K Aug 6 '14 at 3:53
• good guy david here :D – mathsguy Aug 6 '14 at 3:55
• @mathsguy We're right about 270 then? – BCLC Aug 6 '14 at 4:27