How many numbers formed from all the digits 1113333344455678 are there if the digits 6 and 8 have to appear from the different sides of the digit 7?My thoughts:I thought I should first find how many numbers are formed from all the 14 digits 11133333444557, which is ( 14 3 ) ( 11 5 ) ( 6 3 ) ( 3 2 ) = 14 ! 3 ! 5 ! 3 ! 2 ! ,,and, then, analyze the possibilies given a position of the digit 7, but then, I lose the track and I only tell for sure 6 and 8 can be first and last digits in every situation, so there are strictly more numbers than 2 14 ! 3 ! 5 ! 3 ! 2 ! = 14 ! ( 3 ! ) 2 5 ! ..Then, I thought, I should first consider the position p , 1 &lt; p &lt; 16 among 16 overall places for 7 and choose the positions in p − 1 and 16 − p ways for 6 and 8 and multiply the result by 2 for each p, but this didn't lead me anywhere.

Question

How many numbers formed from all the digits 1113333344455678 are there if the digits 6 and 8 have to appear from the different sides of the digit 7?My thoughts:I thought I should first find how many numbers are formed from all the 14 digits 11133333444557, which is                    (                    14      3                      )                                (                    11      5                      )                                (                    6      3                      )                                (                    3      2                      )              =            14      !              3      !      5      !      3      !      2      !        ,,and, then, analyze the possibilies given a position of the digit 7, but then, I lose the track and I only tell for sure 6 and 8 can be first and last digits in every situation, so there are strictly more numbers than   2            14      !              3      !      5      !      3      !      2      !        =            14      !              (      3      !              )        2            5      !        ..Then, I thought, I should first consider the position   p  ,  1  &amp;lt;  p  &amp;lt;  16 among 16 overall places for 7 and choose the positions in   p  −  1 and   16  −  p ways for 6 and 8 and multiply the result by 2 for each p, but this didn&#039;t lead me anywhere.

Amir Beck · Accepted Answer

Step 1One of the approaches is to see that once three positions are decided for 6,7 and 8, you can arrange them within, in   3  !  =  6 ways but here instead of 6, there are only 2 ways to arrange them as 7 must be in between 6 and 8.Step 2So we first find all possible arrangements of 1113333344455678 and then divide by 3 to get all favorable arrangements.So the answer is                1      3        ⋅                  16        !                    3        !        ⋅        5        !        ⋅        3        !        ⋅        2        !