Quandaries and Queries


I am Kevin Cooke I am a parent of four boys aged 8 to 14 years.
My question is as follows.
My 12 year old has asked me how to figure out the number of four digit combinations possible using 0 thru 9 without repeating a number?
Can you help?


Hi Kevin,

It depends what you mean by combinations. If he is thinking of combinations as in a combination lock where the order is important (that is 2,4,6,5 and 2,4,5,6 are different) then you can proceed this way.

Think of constructing the 4 digit combination, reading from left to right. For the fist digit you have 10 choices. Once you have decided on the first digit you have 9 choices for the second digit since you can't repeat digits. Thus for a two digits arrangements you have

10 9 = 90 possibilities.

Now that you have used two digits you have 8 choices for the third digit and hence

10 9 8 = 720 possibilities.

Finally there are

10 9 8 7 = 5040 possible four digit arrangement if order is important.

Mathematicians however use the word combinations for arrangements where order is unimportant. That is (2,4,6,5); (2,4,5,6); (2,5,4,6); (6,4,2,5);... are all the same combination of the four digits 2, 4, 5 and 6. In fact, using the same development as above you can see that there are

4 3 2 1 = 24

different orders of the digits 2, 4, 5 and 6. Hence in the list of 5040 arrangements above every four digit combination is repeated 24 times. With this meaning of combination there are

5040/24 = 210 combinations of the digits 0 thru 9.

There is an expression for this using the factorial notation. For a positive integer n, n factorial, written n!, is defined by

n! = n (n-1) (n-2) ... 2 1

So, for example

4! = 4 3 2 1 = 24

With this notation the number of combinations of 10 digits taken 4 at a time is

 10!/(4! 6!) = 210

In general the number of combinations of n things taken k at a time is

 n!/(k! (n-k)!)



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