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

$(}\genfrac{}{}{0ex}{}{14}{3}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{11}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{6}{3}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{3}{2}{\textstyle )}=\frac{14!}{3!5!3!2!},$,

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 $2\frac{14!}{3!5!3!2!}=\frac{14!}{(3!{)}^{2}5!}.$.

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.

My thoughts:

I thought I should first find how many numbers are formed from all the 14 digits 11133333444557, which is

$(}\genfrac{}{}{0ex}{}{14}{3}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{11}{5}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{6}{3}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{3}{2}{\textstyle )}=\frac{14!}{3!5!3!2!},$,

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 $2\frac{14!}{3!5!3!2!}=\frac{14!}{(3!{)}^{2}5!}.$.

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

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 $\text{}{\displaystyle \frac{1}{3}\cdot \frac{16!}{3!\cdot 5!\cdot 3!\cdot 2!}}$

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 $\text{}{\displaystyle \frac{1}{3}\cdot \frac{16!}{3!\cdot 5!\cdot 3!\cdot 2!}}$

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