Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem11

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11. Let a sub n represent the set of n-digit numbers that do not contain the digit 0. What should the value of n be so that there are as many numbers without the digit 9 in the  set as there are numbers with exactly one digit equal to 9? First let's count the number of elements in the set AN. For each digit we have 9 choices.  So this tells us that AN contains 9 times 9 times 9, where each digit can be chosen in 9 different ways. There are n factors in this product, so that's 9 to the power  n elements. Now, if we want no nines, what we would do is perform the same calculation but each digit can be chosen in 8 different ways. So that's 8 to the power n elements.  And if we want just one 9, then we have n ways to place this 9 and n-1 digits remaining. And each of those n-1 digits can be anything except a 0 or a 9, so that's 8 choices again.  So we have a total of n times 8 to the power n-1 elements. And then what we want is for these to be equal, and that happens if and only if 8 to the power n is equal to n times  8 to the power n-1, and we would like the exponents to be the same. So if n is equal to 8, we have 8 to the power n equal to 8 times 8 to the n-1, and then in fact these  would be equal. That doesn't happen with any of the other choices presented to us, so it must in fact be the case that n is equal to 8, and the answer to this question is A.

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