標題:
math11
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發問:
How many seven-digit numbers contain the digit seven at least once?
最佳解答:
The answer is 4,748,472 FIrstly we need to find the total no. of seven-digit numbers They are (1,000,000 to 9,999,999), Total no. of 7-digit no. = 9,999,999 - 1,000,000 + 1 = 9,000,000 Then we find the no. of seven-digit numbers that does not contain any digit seven Let the no. be A,BCD,EFG A can be any no. from 1 to 9 except 7, i.e. (1,2,3,4,5,6,8,9); we have 8 choices For B to G, they can be any no. from 0 to 9 except 7, i.e. (1,2,3,4,5,6,8,9,0); we have 9 choices, Therefore no. of seven-digit numbers which contains no digit 7 = 8 x 9 x 9 x 9 x 9 x 9 x 9 = 8 x 9^6 = 4,251,528 No. of seven-digit numbers contain the digit seven at least once = Total no. of seven-digit numbers - No. of seven-digit numbers that do not contain any digit seven = 9,000,000 - 4,251,828 = 4,748,472 You may try to test the method with two-digit numbers, which is simple to understand. Total no. of two-digit numbers They are (10 to 99), Total no. of two-digit no. = 99 - 10 + 1 = 90 No. two-digit numbers that does not contain any digit seven Let the no. be AB A can be any no. from 1 to 9 except 7, i.e. (1,2,3,4,5,6,8,9); we have 8 choices For B, it can be any no. from 0 to 9 except 7, i.e. (1,2,3,4,5,6,8,9,0); we have 9 choices, Therefore no. of two-digit numbers which contains no digit 7 = 8 x 9 = 72 No. of two-digit numbers contain the digit seven at least once = Total no. of two-digit numbers - No. of two-digit numbers that do not contain any digit seven = 90 - 72 = 18 They are 17,27,37,47,57,67,77,87 and 97 ----- (9 nos.), and 70,71,72,73,74,75,76,77,78,79 ----(10 nos., but 77 has already been counted before, so also 9 nos.) In total 18 two-digit numbers contains at the digit seven at least once. Hope you get it.
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