 Quandaries and Queries Hi 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?   Regards Kevin 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)!) Penny Go to Math Central