MGMAT, Word Translation, 4th ed., Chapter 4

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MGMAT, Word Translation, 4th ed., Chapter 4

by futjim » Tue Jul 13, 2010 5:50 pm
Dear all,

I posted this on mgmat forum to no avail.

I have a question about a question on Combinatroics Strategy in the textbook.

On pg. 68, there is a sample question:
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 explanation states that I should think of solving the question this way:
a - first digit, only odd
e - last digit, only odd

a b c d e
5 4
5 8 7 6 4
so, the total number of lock codes is therefore 5x8x7x6x4=6720

My question, by automatically assigning the e as 4, aren't we falsely assuming that d will not be an odd number? for example if a = 1, then e can only be e=3, 5, 7, 9. But how can we be so certain that d doesn't contain 3, 5, 7, 9? the only thing I can think of is that the five digit selection process begins with a, then e, then b, c, and d. That way, e can be 4. otherwise, e can't be 4.

Please help me clarify this.
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by Rahul@gurome » Tue Jul 13, 2010 8:45 pm
Let the 5-digit lock code is abcde.
First and last digits can take the values from 1, 3, 5, 7 or 9.
2nd, 3rd and 4th places can take any digit from 0 to 9.
We can fill the first place, "a" in 5 ways
We can fill the 2nd place, "b" in 8 ways (1 digit is used in "a")
We can fill the 3rd place, "c" in 7 ways (2 digits are used in "a" and "b")
We can fill the 4th place, "d" in 6 ways (3 digits are used in "a", "b" and "c")
We can fill the 5th place, "e" in 4 ways (1 odd digit is used in "a", so remaining are 4 digits)

Hence, total number of lock codes = 5x8x7x6x4 = 6720
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