MHB With how many ways can the digits be arranged?

[evinda]

Hey! (Blush)

With how many ways can the digits $1,2,3,4,5$ be arranged?And with how many ways can these digits $1,1,3,4,5$ be arranged?

The digits $1,2,3,4,5$ can be arranged with $5!$ ways,right?

For the second subquestion,I tried to arrange the digits: $1,1,3$ and found that there are $\binom{3}{2}=3$ ways,so is it in our case $\binom{5}{2}$,or am I wrong??

 

[evinda]

I thought about it again.. (Thinking) Is it maybe $\frac{5!}{2!}$ ? Because,we choose firstly $2$ positions for the $1$s.There are $\binom{5}{2}$ possibilities,and then we arrange the left digits at the $3$ left positions.As the digits are discrete,there are $3!$ ways to arrange them.So,there are,in total, $\binom{5}{2} \cdot 3!$ ways to arrange the digits.

 

[Fermat1]

I thought about it again.. (Thinking) Is it maybe $\frac{5!}{2!}$ ? Because,we choose firstly $2$ positions for the $1$s.There are $\binom{5}{2}$ possibilities,and then we arrange the left digits at the $3$ left positions.As the digits are discrete,there are $3!$ ways to arrange them.So,there are,in total, $\binom{5}{2} \cdot 3!$ ways to arrange the digits.

First pretend all possibilities are distinct: so that would $5!$ possibilities. But for each arragnement, 2 are not distinct. So there are $\frac{5!}{2}$ possible arrangements.

 

[evinda]

First pretend all possibilities are distinct: so that would $5!$ possibilities. But for each arragnement, 2 are not distinct. So there are $\frac{5!}{2}$ possible arrangements.

So you mean that the denominator should't be $2!$ ? For example,if we would have to arrange the digits $1,1,1,2,2,3,4,5$ wouldn't there be $\frac{8!}{3!2!}$ ways?

 

[Fermat1]

So you mean that the denominator should't be $2!$ ? For eaxmple,if we would have to arrange the digits $1,1,1,2,2,3,4,5$ wouldn't there be $\frac{8!}{3!2!}$ ways?

you are right,but observe 2!=2

 

[evinda]

you are right,but observe 2!=2

Ok,thank you! :)