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How many 10 digit numbers contain no repetition of digits? - Answers

This is a permutation problem. The digits are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The 1st digit has 9 possibilities (it cannot be zero). Thus, there are 9(9P9) = 9(9!) = 9(362,880) = 3,265,920 such numbers, because: The 2nddigit has 9 possibilities (because we are left with 9 digits). The 3rd digit has 8 possibilities. The 4th digit has 7 possibilities. The 5thdigit has 6 possibilities. The 6thdigit has 5 possibilities. The 7thdigit has 4 possibilities. The 8thdigit has 3 possibilities. The 9thdigit has 2 possibilities. The 10thdigit has 1 possibility. so that, 9*9*8*7*6*5*4*3*2*1= 3,265,920.



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How many 10 digit numbers contain no repetition of digits? - Answers

https://math.answers.com/math-and-arithmetic/How_many_10_digit_numbers_contain_no_repetition_of_digits

This is a permutation problem. The digits are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The 1st digit has 9 possibilities (it cannot be zero). Thus, there are 9(9P9) = 9(9!) = 9(362,880) = 3,265,920 such numbers, because: The 2nddigit has 9 possibilities (because we are left with 9 digits). The 3rd digit has 8 possibilities. The 4th digit has 7 possibilities. The 5thdigit has 6 possibilities. The 6thdigit has 5 possibilities. The 7thdigit has 4 possibilities. The 8thdigit has 3 possibilities. The 9thdigit has 2 possibilities. The 10thdigit has 1 possibility. so that, 9*9*8*7*6*5*4*3*2*1= 3,265,920.



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https://math.answers.com/math-and-arithmetic/How_many_10_digit_numbers_contain_no_repetition_of_digits

How many 10 digit numbers contain no repetition of digits? - Answers

This is a permutation problem. The digits are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The 1st digit has 9 possibilities (it cannot be zero). Thus, there are 9(9P9) = 9(9!) = 9(362,880) = 3,265,920 such numbers, because: The 2nddigit has 9 possibilities (because we are left with 9 digits). The 3rd digit has 8 possibilities. The 4th digit has 7 possibilities. The 5thdigit has 6 possibilities. The 6thdigit has 5 possibilities. The 7thdigit has 4 possibilities. The 8thdigit has 3 possibilities. The 9thdigit has 2 possibilities. The 10thdigit has 1 possibility. so that, 9*9*8*7*6*5*4*3*2*1= 3,265,920.

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      This is a permutation problem. The digits are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The 1st digit has 9 possibilities (it cannot be zero). Thus, there are 9(9P9) = 9(9!) = 9(362,880) = 3,265,920 such numbers, because: The 2nddigit has 9 possibilities (because we are left with 9 digits). The 3rd digit has 8 possibilities. The 4th digit has 7 possibilities. The 5thdigit has 6 possibilities. The 6thdigit has 5 possibilities. The 7thdigit has 4 possibilities. The 8thdigit has 3 possibilities. The 9thdigit has 2 possibilities. The 10thdigit has 1 possibility. so that, 9*9*8*7*6*5*4*3*2*1= 3,265,920.

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