Permutation/Combination Question

[peblez]

Answer the following questions using the letters in" assessment"

How many ways can the letters be arranged if the arrangement must start or end with the letter "s".

so i know this is mutally exclusive event, so i have to add the # of ways the letters can with s and the # of ways it must end with s.

but how do i find those #'s?

 

[Denis]

Try this:

assessment = sssseeamnt = 1111223456

How many 10digit numbers containing these digits will start or end with 1?

In increasing order:
1111223456
1112123456
...
6543221111

 

[pka]

Answer the following questions using the letters in" assessment"
How many ways can the letters be arranged if the arrangement must start or end with the letter "s".

There are \(\displaystyle \frac{{10!}}{{\left( {2!} \right)\left( {3!} \right)}}\) arrangements that begin with an s, there are \(\displaystyle \frac{{10!}}{{\left( {2!} \right)\left( {3!} \right)}}\) arrangements that end with an s, there are \(\displaystyle \frac{{10!}}{{\left( {2!} \right)\left( {2!} \right)}}\) arrangements that begin and end with an s.

Thus the total is:
\(\displaystyle \L \frac{{10!}}{{\left( {2!} \right)\left( {3!} \right)}} + \frac{{10!}}{{\left( {2!} \right)\left( {3!} \right)}} - \frac{{10!}}{{\left( {2!} \right)\left( {2!} \right)}}\)

 

[grapz]

may i ask how you got to the value of 10 ! / 2 ! x 3!, for arrangements that begin with s and end with s?

i am also interested. why isn't it 9! / 2! x 3!?

 

[pka]

YES, you are correct!
The answer should be: \(\displaystyle \L \frac{{9!}}{{\left( {3!} \right)\left( {2!} \right)}} + \frac{{9!}}{{\left( {3!} \right)\left( {2!} \right)}} - \frac{{8!}}{{\left( {2!} \right)\left( {2!} \right)}}\)