How many 3 digit numbers can you form by using the digits 7 8 1 only once?
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Six-digit PINs not more secure than four-digit ones "Mathematically speaking, there is a huge difference, of course," says Philipp Markert. A four-digit PIN can be used to create 10,000 different combinations, while a six-digit PIN can be used to create one million.
I think the analogy with the permutations of letters is making this problem more complicated than it needs to be. Show
Using the restriction that the number has at least one seven, you can first find the numbers that have exactly one $7$, then the numbers that have two $7$s, and then the number that has three $7$s and then add the results. To find the number of numbers, think of choosing a digit for each spot: _ _ _ For one seven, you can fix a $7$ in a spot, say the first one, so the number looks like 7_ _ and note that for the other spots you can have any of the other 9 digits ($0,1,2,3,4,5,6,8,$ or $9$), so there are $9\cdot 9=81$ such numbers. For numbers of the forms _ 7 _ and _ _ 7 the count is different because the first digit cannot be $0$, so there are $8\cdot 9=72$ possibilities for each. Thus, in total there are $72+72+81=225$ three-digit positive integers with one seven as a digit. Two sevens: For the form _77 there are 8 possibilities because the first spot cannot be 0, and for each of the forms 7_7 and _ _7 there are 9 possibilities, so in total there are $8+9+9=26$ three-digit positive integers with one seven as a digit. Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?Solution:
Problem 3: How many 4-letter code can be formed using the first 10 letters of the English alphabet if no letter can be repeated?Solution:
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