Expert: Josh - 6/28/2006

Question
I am about to take the GMAT and have not had math in quite some time.

My study quide presented a question which I thought I knew how to answer but after seeing the answer I understand that mine was wrong.

I don't remember basics about combination possibilities.  Nor do I remember the difference between 3N and 3!.

If a question was given regarding the potential combinations of a lock with 3 dials, each having the numbers 0 - 9,  I remember that to figure this it has something to do with 9*8*7*6*5*4*3*2*1 = ? or am I way off on this?

The sample question I had in a prep book was very similar to this.

Thanks.

Answer
Hi Bryant,

You're on the right track.
The thing you referred to (n!) is called a factorial.

By definition, n!=n*(n-1)*(n-2)*...*1. So, if n=5, then, 5!=5*4*3*2*1.

It's useful to remember the first few.
0!=1 by definition.
1!=1
2!=2*1=2
3!=3*2*1=6
4!=4*3*2*1=24
5!=120
6!=720
7!=5040
and so forth.

(A) Understanding the concept of factorial will put you in good stead. We normally associate n! with the number of ways of selecting n objects, given n objects. Each time, the object is removed after selection, without replacement. Here, we label each object with numbers for simplicity.

e.g., If n=4, the first selection from the set {1,2,3,4} leads to four distinct possibilities. I can get a "1", "2", "3" or "4". Suppose that I picked "2" for the sake of argument. Then, I'm left with {1,3,4} since "2" is now removed. The second selection leads to three distinct possibilities. Suppose that I picked "3" this time, I'll be left with {1,4}. For the third selection, I have only two choices. If I picked "1", I'll be left with {4}. The fourth and final selection produces only one possibility. I'll definitely get a "4". So, multiplying these contributions, we get 4*3*2*1 = 4! = 24 different possible outcomes. This is equivalent to asking "in how many ways can I rearrange n objects".

Note: Something you may want to look up on is "permutation" and "combination". The former, P(N,k)=N!/(N-k)! concerns with the number of ways of selecting k objects from N objects, where the order of selection matters. The latter, C(N,k)=N!/[k!*(n-k)!] concerns with the number of ways of selecting k objects from N objects, where the order of selection does NOT matter; hence the word "combination".

(B) Getting back to your question. We have 3 dials which may be set independently to produce a combination code.

The first dial contributes to 9 possible different outcomes. Likewise, the second and third dial each also contribute to 9 possible outcomes (as there are 9 single digits to select from). Because each digit may be chosen independently, we apply the "multiplication rule" and argue that they produce 9*9*9 possible outcomes. (No factorial is used here)

Cheers.