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 Post subject: Maths Help - chocolate!
PostPosted: Tue Feb 26, 2013 10:57 pm 
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How many different word combinations can be made using the letters from the word chocolate? (don't have to be words in the dictionary).

Is there a formula one should use?

EDITED to add: Question set for a YEAR 8 DC!


Last edited by Pumpkin Pie on Tue Mar 05, 2013 12:42 pm, edited 1 time in total.

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PostPosted: Tue Feb 26, 2013 11:28 pm 
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Location: Warwickshire
If all 9 letters were different the number of combinations would simply be:
9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

However there are two c's and 2 o's in chocolate

Each of those halves the number of possibilities.


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PostPosted: Wed Feb 27, 2013 9:25 pm 
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Yes there is.

The question is ambiguous because you don't say if you need to use ALL the letters. If you only want 9 letter words, then the formula is the permutation formula, nPr, divided by repeated elements.

For formula for the number of permutations of r objects picked from n is n!/(n-r)!. I.e. 9!/0! in this case, or just 9! in the case when you use all 9 letters (since 0! = 1).

Where there are repeated elements, we must divided by the number of each repeated element, i.e. 2! and 2!. 2! is the same as 2, so it's just 9!/2/2 (which is the same as what Okanagan says) - there are nine letters, and two groups of two, so you take 9! (factorial) and divide by 2! for each group

If you want all the answers, i.e. one-letter words up to nine-letter words, it is much more complicated:

There are 7 different letters, and two of the letters are repeated twice, so you've got 5 single letters and 2 double letters

Therefore with 8 letters you can have:

2 Cs and an O, or 2 Os and a C, plus all the other letters, which is:
= 8!/2! * 2 (times two for the Cs and O vs Os and C)
or
2 Cs, 2 Os, and 4 of the 5 other letters
= 8!/2!/2! * 5 (times five because there are obviously five identical patterns, each missing one of the undoubled letters)

Which is

(8!/2!)* (2 + 5/2!) = 20160 * 4.5 = 90,720

Which is of course equivalent to 9!/1!/2!/2!, which is identical to 9P8 / 2!/2!

Going down to 7 letters, you have the following patterns:

5 single letters, a C and an O
= 7! * 1 (only one such pattern)
2 Cs or 2 Os, 5 single letters
= 7!/2! * 2 (either 2 Cs or 2 Os)
2 Cs and 2 Os, 3 single letters
= 7!/2!/2! * (5!/3!/2!)
= 7!/2!/2! * 10 [10 possible ways of picking 3 single letters from 5]
2 Cs, 1 O, or vice versa, 4 single letters
=7!/2! * 2 * 5 [as before, 5 possible missing letters]

Which is

7! * (1 + 2/2! + 10/2!/2! + 5*2/2!)
=
7! * (1 + 1 + 2.5 + 5)
=
7! * 9.5
= 47,880

This however is larger than 9P7 / 2!/2!, which only comes to 45,360. Obviously the fewer letters you use the less significant the repetition comes, so when you go down to 6, 5, 4 letters, the difference between nPr/2!/2! and the actual number of permutations.

Note that the formula 5!/3!/2! which is from the formula for combinations (nCr), for picking 3 letters from 5.

You could write them out by hand, assuming there are 5 letters, ABCDE, and we must pick 3 of them, we can easily see that there are 10 possible options:

ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE

which is identical to 5!/3!/2!. If you consider combinations/permutations as being based around factorials, then the 5! represents the permutations of 5 letters, and the 3! is the permutations of the 3 letters we are choosing, and the 2! is the permutations of the letters that we aren't.

To get the correct answer to the number of possible words we can make from chocolate, we would have to continue all the way down to 1 letter 'words' (of which there are obviously 7), and then sum the results - 9-letter 'words', plus 8-letter words, plus 7-letter words, etc.


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PostPosted: Wed Feb 27, 2013 10:33 pm 
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Thank you twelveminus for taking the time to give such a detailed answer, and thank you okanagan too, for your answer.

I'm sorry, but I should have mentioned that all 9 letters needed to be used.

So, if I understand correctly, the answer would be:

9x8x7x6x5x4x3x2x1=362880

362880/2=181440

181440/2=90720

Ans= 90720



Just one more point, what does ! mean when placed after a number?

Many thanks!


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PostPosted: Wed Feb 27, 2013 10:38 pm 
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Joined: Mon Aug 22, 2011 8:20 pm
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Location: Warwickshire
Pumpkin Pie wrote:
Just one more point, what does ! mean when placed after a number?
It's a short way of writing the number multiplied by all the other numbers between itself and 1. So:
9! would be 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
8! would be 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
7! would be 7 x 6 x 5 x 4 x 3 x 2 x 1
and so on.


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PostPosted: Wed Feb 27, 2013 10:47 pm 
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7! is shorthand for 7 factorial which is equal to 7 x 6 x 5 x 4 x 3 x 2 x 1

It is not part of the primary curriculum


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PostPosted: Wed Feb 27, 2013 11:02 pm 
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Thank you Guest55, now I can make sense of the detailed explanation! :D


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PostPosted: Tue Mar 05, 2013 12:11 pm 
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Is this question for SATs or 11 plus? I am wondering I should teach my son these questions


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PostPosted: Tue Mar 05, 2013 12:39 pm 
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Karupalli, sorry, I should have said earlier, but this was a maths question DS in YEAR 8 was doing. Nothing to do with the 11+.

Sorry to cause any confusion! :)


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PostPosted: Tue Mar 05, 2013 12:47 pm 
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Interestingly, a factorial question was on a 10+ scholarship paper that my DS sat this year.


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