Combinatorics

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Combinatorics

by Abdulla » Tue Dec 16, 2008 5:51 pm
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?

Can someone explain why 4 x 5 x 8 x 7 x 6 = 6,720
OR if there is another approach please share it.
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by brb588 » Tue Dec 16, 2008 6:22 pm
Hi.

Okay, five digits, two are odd, none can repeat. Let start with the first digit. From 0-9, there are five odd digits (1, 3, 5, 7, 9). The first digit has 5 options.

5 * _ * _ * _ * _

Now let's jump to the fifth digit. If no numbers repeat, there can only be four numbers that digit five can be:

5 * _ * _ * _ * 4

Now fill in the middle digits. Again, no numbers can repeat. Two digits are already filled in, leaving eight for the second digit, seven for the third digit, and six for the fourth.

5 * 8 * 7 * 6 * 4 = 6720

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by Abdulla » Tue Dec 16, 2008 6:38 pm
Thanks , got you.. but do you have another approach .. I mean by using the permutation formula??
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by Brent@GMATPrepNow » Tue Dec 16, 2008 6:48 pm
The permutation formula is only used when you have n objects and you want to arrange some of those objects.
For example, you have 10 people and you want to seat (arrange) 4 of them in a row.

In the original question, we don't have that sort of situation.

In my opinion, there are few occasions when you have to use permutations on the GMAT. The Fundamental Counting Principle will do just fine.

The combinations formula, on the other hand, is definite must, especially if you are scoring in the 600+ range
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by vittalgmat » Wed Dec 17, 2008 1:57 am
Abdulla wrote:Thanks , got you.. but do you have another approach .. I mean by using the permutation formula??
IMO, the counting method with a pictorial representation, as done by brb88 is easier.

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