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 Post subject: digits
PostPosted: Sun Jun 08, 2008 3:40 pm 
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Joined: Fri Mar 07, 2008 9:27 am
Posts: 18
Hi
anyone know this answer, could you please explain this answer?

These are the four two-digit numbers that can be made with the digits 1and 7.

11 17 71 77

How many three-digit numbers can be made with the digits 1,2 and 3?.

(a) 6
(b) 9
(c) 21
(d) 27


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 Post subject:
PostPosted: Sun Jun 08, 2008 3:43 pm 
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Joined: Mon Feb 12, 2007 1:21 pm
Posts: 11934
27 I think

each time there are three numbers to chose from so 3 x 3 x 3


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PostPosted: Sun Jun 08, 2008 7:33 pm 
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Joined: Fri Mar 17, 2006 5:12 pm
Posts: 1300
Location: Birmingham
Hi Lala

27 is incorrect

The question is asking for the number of permutatons of 3 numbers taken three at a time.

There is a matherical formula which is: p = n!/ (n-r)!

Where n! means factorial = n x (n-1) x (n-2).....x 1

Hence for this p = 3 x2 x1 = 6

You can also work this ourt using the various sequences:-

Start with 1 in the Hundreds column: there are 2 options 123 and 132

With 2 you have 213 and 231

and with 3 you have 312 and 321

A total of 6.

Ken


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PostPosted: Sun Jun 08, 2008 7:55 pm 
Dear Ken,

If you read the question (as we're always telling our kids!) you seem to be able to use the digits more than once. So you can have the 27 I think.

e.g.

111, 112, 121, 122, 113, 131, 133, 123, 132,

222, 212, 211, 221, 213,231, 233, 223, 232,

333, 311, 312, 321, 322, 331, 332, 323, 313

Something like that.

Regards,

Fm


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PostPosted: Sun Jun 08, 2008 8:32 pm 
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Joined: Fri Mar 17, 2006 5:12 pm
Posts: 1300
Location: Birmingham
Hi fm

You are quite correct - didn't spot you can reuse the digits


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