# How many 10 letter 'words' can be made with no repeated letters and at least 8 consonants?

Need a little bit of help with this one.

How many 10 letter 'words' can be constructed which have no repeated letters and at least 8 consonants?

where 'word' is defined as any combination of the 26 letters of the english alphabet, and by consonant, (obviously) a letter of the english alphabet that are not either a,i,u,e, or o.

This is where I am at:

• We need to choose the positions for the 8 consonants which is $c(10,8)$

• Then we need to choose the 8 consonants which I think is $p(21,8)$? (not sure about this one)

• Choose the remaining 2 letters: We have already used 8 letters so we only have 18 letters to choose from so $p(18, 2)$

What else needs to be calculated? Do we need to choose the position of the remaining 2 letters?

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Note to everyone: this is similar to math.stackexchange.com/questions/39428/… , but it's different enough that I think it warrants its own solution. – El'endia Starman May 18 '11 at 3:31

Your argument has a couple of problems:

• You don't specify where the last two letters selected go; you pick the two letters, but you never place them; placing them in different order gives you different solution.

• Likewise, you never say how you will distribute the 8 consonants you pick; you select the 8 locations, but you don't say which letter goes where.

• Finally, your solution counts some words too many times. Say you first select the eight consonants b,c,d,f,g,h,j,k and the first eight positions for them (in that order). Then you select two more letters, say m,n, and place them in order in positions nine and ten. You will "count" this word again if your eight selected consonants are b,c,d,f,g,h,m,n, the eight positions selected are positions one through six, nine, and ten (in that order); and then the other two letters you select are j and k and you place them in positions seven and eight. You will count this word several times through your process.

So: perhaps it's better if we start by counting how many 10 letter strings have exactly eight consonants and no repeated letters. My recommendation:

• First select the eight positions that will contain the consonants.
• Then select the consonants and place them in those positions in order (that is, which consonant you select first matters, because that will go in the leftmost selected position).
• Finally, select the two vowels that will fill the remaining two positions, again in order.

Then count how many of your strings have exactly 9 consonants in a similar manner; then how many have all letters consonants (that will be the simplest one).

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Why do I need to select two vowels? The remaining two spots can be any letters, not necessarily vowels since the question says "At Least" 8 vowels – Arvin May 18 '11 at 7:43
@Arvin: You want to count the number of words with at least 8 consonants. I'm suggesting you break this up into three parts: the number of words with exactly 8 consonants, the number of words with exactly 9 consonants, and the number of words with exactly 10 consonants. Taken together, this will be the whole set. I specifically said that we were counting the words that had exactly 8 consonants, not "at least" 8 consonants. So, for this part of the count, the remaining two positions must be vowels. – Arturo Magidin May 18 '11 at 15:30
Ahhh, I see, I see, yeah, for a 10 letter word to have at least 8 consonants it means it either has exactly 8 consonants, or it has exactly 9 consonants or it has exactly 10 consonants! Thanks! – Arvin May 19 '11 at 5:05

I think the best way would be to separate it into 3 non-overlapped groups:

1. Words with exactly 8 consonants
2. Words with exactly 9 consonants
3. Words with exactly 10 consonants

Logic would be similar with Arturo. Arvin comments wouldn't work, because you count same words several time. Once when the consonant is one of the 8, and again, when it happens to be one of the two remaining letters.

For the 8 consonants:

• Select 8 positions - C(10,8)
• Select the 8 consonants out of the 21, in order - P(21,8)
• Select the 2 vowels in order - P(5,2)

So answer is: c(10,8) * p(21,8) * p(5,2) = (10!/(2!*8!)) * (21!/13!) * (5!/3!) = 45 * (21!/13!) * 20

Similarly, do the same for 9 and 10:

• 9 consonants: C(10,9) * P(21,9) * P(5,1) = 10 * (21!/12!) * 5 = 50 * (21!/12!)
• 10 consonants: C(10,10) * P(21,10) * P(5,0) = (21!/11!)

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In the "8 consonants" part, the third bulleted item should be "Select the 2 vowels". – Arturo Magidin May 18 '11 at 15:28
Yes I see, this is the same solution as Arturo's but in mathematical notation. Thanks for this, now I can do the question myself from the start and compare it with yours! – Arvin May 19 '11 at 5:08
@Arturo, Are you guys sure we are meant to be using Permutations (as in p(21,8) and p(5,2))? The question asks for No Repeated letters. Shouldn't we be using Combinations? – Arvin May 20 '11 at 12:04
@Danra, thank you for the edit! – Uri Dec 18 '12 at 16:59

The problem with your approach is that words of 10 consonants will get counted many times. You can pick any two places for the "not necessarily consonants" and finish with the same word. The approach suggested by user6312 in your previous post will work here, too.

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Ross, what do you mean the words of 10 consonants will get counted many times? Words with 10 consonants do not "Include" words with 9 consonants, or 8 consonants, as they are treated as different Words – Arvin May 19 '11 at 5:07
@Arvin: take a word with 10 consonants. You can get it by picking positions 1-8 in step 1, filling them with consonants in step 2, then filling 9 and 10 with consonants in step 3. You can also get it by picking positions 3-10 in step 1 and so on. So it will be counted $\binom {10}{2}=45$ times. This is the advantage of counting 8,9, and 10 consonant words separately. For an 8 consonant word, you force the last two to be vowels and don't overcount. – Ross Millikan May 19 '11 at 12:52

The number of 10 letter words for any given 10 unique letters is simply 10!. Once you notice that the problem becomes how many unique sets of 10 letters are there for the problem at hand.

Breaking it down:

1. Calculate the number of unique sets of ten letters with 8-10 consonants and 0-2 vowels.
2. Multiply by the number of position permutations for a list with 10 items (10!).

The first step has three cases that need to be added together: words with 0, 1 or 2 vowels. So 21*20*19*18*17*16*15*14*13*12 + 5*21*20*19*18*17*16*15*14*13 + 5*4*21*20*19*18*17*16*15*14 = 1977336748800. Multiply that by 10! and you get: 7,175,359,594,045,440,000.

Or in your notation: (p(21,10) + p(5,1)*p(21,9) + p(5,2)*p(21,8)) * p(10)

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