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digits
Posted: Sun Jun 08, 2008 4:40 pm
by lala
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
Posted: Sun Jun 08, 2008 4:43 pm
by Guest55
27 I think
each time there are three numbers to chose from so 3 x 3 x 3
Posted: Sun Jun 08, 2008 8:33 pm
by KenR
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
Posted: Sun Jun 08, 2008 8:55 pm
by fm
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
Posted: Sun Jun 08, 2008 9:32 pm
by KenR
Hi fm
You are quite correct - didn't spot you can reuse the digits