Dear all,

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.

futjim,
The question says that only first and last digit needs to be. However the numbers in position: can either be odd or even.

A B C D E
O O/E O/E O/E O



Let me explain further with an example that you picked up

A B C D E
1 - - - 3

Now D has choice of 2-4-5-6-7-8-9-0 ==> 8 choices
C will have 7 choice (after we pick a number for D)
B will have 6 choice (after we pick a number for C)

It does not matter what number is picked for B,C and D. You can take any other example and count how many number of choices does D have and so on.

I hope it makes things clear.

Thank you so much for answering as I've posted the same question up on other boards and I haven't been getting much answers for this question. I'm probably not very articulate with this question and hence the lack of response.

Getting back to your example


I agree that the numbers picked for B,C, and D doesn't matter but what about the order (which is the heart of the question)

1st draw: A = 1
2nd draw: E = 3

The question never states what the order of the number generation process is, which is confusing the heck out of me. If E was drawn 2nd then your explanation is correct but what if E was drawn last? What if B,C,D were all odd numbers? then E's choices would be reduced because of the 'no reptition' rule, right?

Maybe I'm just being pedantic.

If you or anyone else can, please help!

Ok. Let's come to what is confusing you.

1. "The question never states what the order of the number generation process is, which is confusing.." - True. The question does not say the order. The question say that first and last ARE odd. It does not matter what odd number are chosen or in what order. YOU MUST keep aside 2 odd numbers and fill the remaining 3 places with 8 numbers.

2. "What if B,C,D were all odd numbers? then E's choices would be reduced because of the 'no repetition' rule...." ----- Yes BCD CAN be odd, but OUT OF HOW MANY CHOICES. B - lets' pick odd number OUT OF 8 numbers i.e. 5 even + 3 odd (as two are already fixed for A and E position). After pick a number for place B, pick a number for place C OUT OF 7 numbers i.e 4 Even + 3 odd (if you picked even for B) OR 5 Even + 2 odd (if you picked odd number for B) and so on.

I hope the explanation helps.

Thanks dinesh19aug,

I totally understand now.

thanks, dinesh..

_________________
Tim Sanders
Manhattan GMAT Instructor

Dear GMAT Instructor,

For the question stated above, these were my deductions and how I came about the answer:

Sample Set of numbers to choose from:
{ 0,1,2,3,4,5,6,7,8,9}

Odd Numbers: 1,3,5,7,9, Therefore total 5 odd numbers
Total Numbers in the set: 10

First Number: ODD / 5 possibilities
Last Number: ODD / 4 possibilities
2nd Number: 10-2 = 8 possibilities
3rd Number: 10-3 = 7 possibilities
4th Number: 10-4= 6 possibilities

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

Does this sound right or am I totally missing the point here?

sirat, perfect! I completely agree with your thought process.

_________________
Jamie Nelson
ManhattanGMAT Instructor