MGMAT Strategy Guide Question

[sunman]

From Number Properties Strategy Guide, Page 48 - Combinatorics.

An Office Manager must choose a five digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?

The book answer says:

5 x 8 x 7 x 6 x 4 = 6720.

I don't see it.

I went with:

5 (5 odd numbers) x 9 (9 numbers remaining, including zero) x 8 x 7 x ???

I got stuck at the end, because we don't know whether or not some of the odd numbers have already been used up in the middle (the question says the digits at the end have to be odd, but it doesn't say the digits in the middle MUST be even!), so the final digit could be 4, but it also could be 1, depending on how many digits in the middle were odd.

Experts, please assist? Is the book wrong, or am I lost in the sauce?

Much appreciated,

Sonny

[Anurag@Gurome]

5 (5 odd numbers) x 9 (9 numbers remaining, including zero) x 8 x 7 x ???

I got stuck at the end, because we don't know whether or not some of the odd numbers have already been used up in the middle (the question says the digits at the end have to be odd, but it doesn't say the digits in the middle MUST be even!), so the final digit could be 4, but it also could be 1, depending on how many digits in the middle were odd.

That's why the strategy for solving counting problems is to satisfy the restrictions first then select freely. In this case we have to fill up the first and last place with odd digits first. Then we can freely select digits for middle three places.

An Office Manager must choose a five digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?

First place can be filled in 5 ways as there are 5 odd digits.
Last place can be filled in 4 ways as there are 4 odd digits left.
Second place can be filled in 8 ways as there are (10 - 2) = 8 digits left.
Third place can be filled in 7 ways as there are (10 - 3) = 7 digits left.
Fourth place can be filled in 6 ways as there are (10 - 4) = 6 digits left.

Hence, total number of possible codes = 5*4*8*7*6 = 6720

[sunman]

Hmmm...ok so first and last need to be filled first.