Distinct two-digit numbers are written one after the other to form a six-digit number. How many six-digit numbers thus formed have four consecutive 1s in them?
Question
Distinct two-digit numbers are written one after the other to form a six-digit number. How many six-digit numbers thus formed have four consecutive 1s in them?
Solution
To solve this problem, we need to consider the different positions the four consecutive 1s can be in the six-digit number. There are three possible cases:
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The 1s are in the first four positions: 1111XY. There are 90 choices for XY (10-99), so there are 90 such numbers.
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The 1s are in the last four positions: XY1111. There are 90 choices for XY (10-99), so there are 90 such numbers.
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The 1s are in the middle four positions: X1111Y. There are 9 choices for X (1-9) and 10 choices for Y (0-9), so there are 9*10 = 90 such numbers.
Adding up these cases, there are 90 + 90 + 90 = 270 six-digit numbers that have four consecutive 1s.
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