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How many permutations exist of the letters a b c d taken four at a time?
Since you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.
How many license plates are possible consisting of 4 letters followed by 3 digit if the first letter must be a C?
1x26x26x26x10x10x10=17576000
How many distinguishable permutations of letters are possible in the word critics?
The word critics has 7 letters which can be arranged in 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 ways. To see this, imagine placing one of the 7 letters in the first slot. Then place a different letter in the second slot (there are only 6 letters left now), then 5 and so on down to 1. We multiply because we must do each of these steps to create a rearrangement of the word critics. The problem then is that critics contains 2 'c's and 2 'i's, which are indistinguishable. For example, we might count "rtiicsc" several times by switching the places of the 'i's or the 'c's even though we cannot tell the difference in the word. So, we must divide out the repetition, by dividing by 2! = 2 * 1 twice. This corrects the over-counting from the duplicate letters. So the correct result is 7!/(2! * 2!) = 1260 distinguishable permutations.
How many permutations of 3 different digits are there from the ten digits 0 to 9 inclusive?
ABC A = 1,2,3,4,5,6,7,8,9; 9 possibilities B = 0,1,2,3,4,5,6,7,8,9; 10 possibilities C = 0,1,2,3,4,5,6,7,8,9; 10 possibilities So there are 9*10*10 = 900 numbers with 3 digits.
How many 6 digit numbers can be formed with 4 different digits?
The question can be re-stated as asking for the total number of permutations that can be derived from the following two groups of digits: AAABCD and AABBCD, where A, B, C and D are different. The number of ways of choosing the digit A, to be used three times, out of the ten digits {0, 1, 2, ... 9} is 10. Having done that, the number of ways of selecting 3 from the remaining 9 digits is 9C3 = (9*8*7)/(3*2*1) = 84. Thus there are 10*84 = 840 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 3 of the digits are the same, these permutations are not all distinct. In fact, there are 6!/3! = 120 distinct permutations. That makes a total of 840*120 = 100800 such numbers. Next, the number of ways of choosing the digits A and B, each to be used twice, out of the ten digits {0, 1, 2, ... 9} is 10C2 = (10*9)/(2*1) = 45. Having done that, the number of ways of selecting C and D from the remaining 8 digits is 8C2 = (8*7)/(2*1) = 28. Thus there are 45*28 = 1260 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 2 pairs of these digits are the same, these permutations are not all distinct. In fact, there are 6!/(2!*2!) = 180 distinct permutations. That makes a total of 1260*180 = 226800 such numbers. The grand total is, therefore, 100800 + 226800 = 327600 6-digit numbers made from 4 distinct digits.
