permutation

• Jul 2nd 2013, 12:14 AM
Trefoil2727
permutation
How many 4-digit numbers with values 6000 and above can be formed from the digits 4,5,6,7,8, where every digit cannot be used more than once in a number? Among the numbers formed, how many are odd number?

I've got 3 X 4 X 3 X 2 =72, but I still can't answer the second question..
• Jul 2nd 2013, 02:41 AM
Plato
Re: permutation
Quote:

Originally Posted by Trefoil2727
How many 4-digit numbers with values 6000 and above can be formed from the digits 4,5,6,7,8, where every digit cannot be used more than once in a number? Among the numbers formed, how many are odd number?
I've got 3 X 4 X 3 X 2 =72, but I still can't answer the second question..

There are $(3)(3)(2)(1)$ of those numbers end in $5$.

There are $(2)(3)(2)(1)$ of those numbers end in $7$.

So what is the answer to the second question?
• Jul 3rd 2013, 04:43 AM
Trefoil2727
Re: permutation
Quote:

Originally Posted by Plato
There are $(3)(3)(2)(1)$ of those numbers end in $5$.

There are $(2)(3)(2)(1)$ of those numbers end in $7$.

So what is the answer to the second question?

12, how did you get that?
• Jul 3rd 2013, 05:22 AM
Plato
Re: permutation
Quote:

Originally Posted by Trefoil2727
12, how did you get that?

Any of the four digit numbers described in the question must end in 5 or 7 to be odd.
• Jul 3rd 2013, 05:24 AM
ebaines
Re: permutation
Quote:

Originally Posted by Trefoil2727
How many 4-digit numbers with values 6000 and above ... I've got 3 X 4 X 3 X 2 =72,

Did you mean to say 60,000 and above? If so, your answer is correct. But if in fact the question asks about numbers 6000 and above then any combination of all 5 digits will do it, plus any 4-digit number that starts with 6, 7, or 8.
• Jul 3rd 2013, 05:31 AM
Plato
Re: permutation
Quote:

Originally Posted by ebaines
Did you mean to say 60,000 and above? If so, your answer is correct. But if in fact the question asks about numbers 6000 and above then any combination of all 5 digits will do it, plus any 4-digit number that starts with 6, 7, or 8.

The OP says 4-digit numbers,

Quote:

Originally Posted by Trefoil2727
How many 4-digit numbers with values 6000 and above can be formed from the digits 4,5,6,7,8, where every digit cannot be used more than once in a number?

• Jul 3rd 2013, 05:50 AM
ebaines
Re: permutation
Quote:

Originally Posted by Plato
The OP says 4-digit numbers,

Ah! I should have read the original post moire carefully. Thanks.
• Jul 3rd 2013, 06:31 AM
Trefoil2727
Re: permutation
Quote:

Originally Posted by Plato
Any of the four digit numbers described in the question must end in 5 or 7 to be odd.

can I use (3)(2)(1)(2)? I count 5 and 7 together
• Jul 3rd 2013, 07:58 AM
Plato
Re: permutation
Quote:

Originally Posted by Trefoil2727
can I use (3)(2)(1)(2)? I count 5 and 7 together

No you cannot. If the number ends 5 you can still have three choices for the first digit.

However, if the number ends 7 you have only two choices for the first digit.