



 
Hi Natalie, I assume that by "you can use any of the numbers twice" you mean not 3 of 4 times. I also assume that order is important, that is 7475 and 7745 are different. How many different 4digit combinations are possible if you can use each of the 4 digits, 7, 4 and 5, as many times as you want? In this case there are 3 choices for each of the 4 digits so the answer is 3 × 3 × 3 × 3 = 81. How many of these use one of the digits 3 or 4 times? These have to be removed from the 81 you have. Concentrate on 7 for the moment. How many have either 3 or 4 sevens? There is only one containing 4 sevens, 7777. There are four that contain 3 sevens and a four, 4777, 7477, 7747 and 7774. Can you complete the problem now?  


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