How many numbers formed from all the digits 1113333344455678 are there if the digits 6...

Wronsonia8g

Wronsonia8g

Answered

2022-07-03

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 < p < 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.

Answer & Explanation

Amir Beck

Amir Beck

Expert

2022-07-04Added 13 answers

Step 1
One 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 2
So 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 !

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