If no digit can be used more than once, find how many numbers can be formed from the digits 3,4,5,6,7,8 which are six-digit numbers and even.

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- Sep 21st 2006, 11:05 PMclassicstringsFurther Probability
If no digit can be used more than once, find how many numbers can be formed from the digits 3,4,5,6,7,8 which are six-digit numbers and even.

- Sep 21st 2006, 11:11 PMCaptainBlack
Since you have six digits and are asked for six digit numbers, this is

equivalent to asking how many permutations are there of six distinct

objects.

The number of permutations of six distinct objects is 6!=720.

You can see that this is the case as the first digit in the perm can be

any of the six, the second any of the remaining 5, the third any of the

remaining 4, ... The total number of permutations is the product of

these, that is 6x5x4x3x2x1=6!

RonL - Sep 21st 2006, 11:16 PMclassicstrings
- Sep 21st 2006, 11:34 PMCaptainBlack
Oops, sory about that, I did not read the question carefully enough:o

For the numbers to be even work from the least to the most significant

digit.

The least significant can be any of 3, then the next any of the remaining 5,

and so on as before.

So then we have 3x5x4x3x2x1=360

RonL - Sep 21st 2006, 11:41 PMclassicstrings
:confused: Can you tell me why my logic is incorrect?

I think about it this way, at the start you have 6 digits so it can be any of the six.

Then you are left with 5 you can still choose any of the five.

Then if the firs two were both even you cant choose the last even so the possibilities are now 3.

Then it's 2, then 1, and last you are left with 1 even number.

So 6x5x3x2x1x1. But it's wrong :confused: - Sep 21st 2006, 11:56 PMCaptainBlack
You don't really have the freedom to choose any of the six digits for

the most significant, (and 5 for the next ..), because you must have

an even digit left when you reach the least significant digit. But if you

have been selecting the earlier digits from all of the available digits

you may have no even digits left at the end.

The number of choices later in the sequence available if you place no

constraint on how you choose the first two depend on what you actually

chose for the first two so multiplying the numbers together this way does not work.

Much better to start choosing from the other end, where the constraint

can be dealt with easily.

RonL - Sep 22nd 2006, 12:08 AMclassicstrings
Thanks for the explanation, from now on i'll always eliminate the constraint first, then the rest follows. Cheers.