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What is the number of four digit numbers that can be formed from the digits $0,1,2,3,4,5,6,7$ so that each number contains digit $1$?

The answer is $750$ but I am not able to find it.

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closed as off-topic by uniquesolution, Rohan, Dave, hardmath, Ove Ahlman Jan 18 '18 at 16:33

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  • $\begingroup$ Please complete your sentence, you seem to have cut off there... $\endgroup$ – VortexYT Jan 18 '18 at 13:07
  • $\begingroup$ Are the digits necessarily different? $\endgroup$ – user_194421 Jan 18 '18 at 13:57
  • $\begingroup$ Yes , u cannot use one digit more than once $\endgroup$ – user163054 Jan 18 '18 at 14:06
  • $\begingroup$ There are 4 choices for where the $1$ is located, and of the 3 other digits, the leftmost has 7 choices, the 2nd leftmost has 6 remaining choices, and the rightmost has 5 remaining choices. $\endgroup$ – Paul Sinclair Jan 18 '18 at 16:31
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Since we are looking for $4$-digit numbers, the first (thousands) digit can't be $0$.

Let's split cases:

Case 1: first digit is $1$.

In this case, we pick $3$ digits in order out of the remaining $7$, there are $7\cdot6\cdot5=210$ possibilities.

Case 2: first digit is not $1$.

There are $6$ ways to choose the first digit ($2,3,...,7$), $3$ ways to place the digit $1$ (hundreds, tens, ones), and $6\cdot5$ (pick $2$ in order from the remaining $6$ digits) ways to choose the other digits. So there are $6\cdot3\cdot6\cdot5=540$ possibilities in this case.

Total

In total, there are $210+540=750$ possibilities to choose the number.

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  • $\begingroup$ The answer is 480, I already mentioned that in question. $\endgroup$ – user163054 Jan 18 '18 at 15:09
  • $\begingroup$ @user163054 If it's so, how do you know? $\endgroup$ – user_194421 Jan 18 '18 at 15:52
  • $\begingroup$ I only have the answer, not complete solution $\endgroup$ – user163054 Jan 18 '18 at 15:58
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    $\begingroup$ @user163054 Unless there are additional conditions that you did not mention, the correct answer is $750$. $\endgroup$ – N. F. Taussig Jan 18 '18 at 15:59
  • $\begingroup$ There isn't anything additional. And this question appeared in a test and many students got the answer. $\endgroup$ – user163054 Jan 18 '18 at 16:01

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