# 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, 2014 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. Aug 6, 2014 at 3:24
• Hmmm... but Night's answer seems correct. Aug 6, 2014 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) Aug 6, 2014 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. Aug 6, 2014 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.) Aug 6, 2014 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? Aug 6, 2014 at 3:53
• Thanks, this makes perfect sense now. Aug 6, 2014 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 :/ Aug 6, 2014 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. Aug 6, 2014 at 3:53
• good guy david here :D Aug 6, 2014 at 3:55
• @mathsguy We're right about 270 then?
– BCLC
Aug 6, 2014 at 4:27