1. ## Permutation

Hi everyone,

It seems to be very simple problem but sometimes it happens that despite teaching so many years u got stuck in a very simple question

My problem is:

How many ways we can arrange the word permutations in such a way that there always occur 4 letters between p and s.

plz giveme complete descriptuion with the answer

Thanks a lot.

2. There are 7 possible positions for p and s.Rest of the 10 letters can be arranged in 10!/2! ways(since there are 2 T's). p and s can themselves be arranged in 2 ways(i.e.p and s and then s and p)
2.7.(10!/2!)
where do you teach

3. Originally Posted by shivani
Hi everyone,

It seems to be very simple problem but sometimes it happens that despite teaching so many years u got stuck in a very simple question

My problem is:

How many ways we can arrange the word permutations in such a way that there always occur 4 letters between p and s.

plz giveme complete descriptuion with the answer

Thanks a lot.
shavni,

First ask how many ways you can place the p and s. The leftmost of the two can be in position 1, 2, ..., or 7, and once the leftmost is placed the location of the other letter is fixed. (For example, if the p is leftmost and is in location 7, then the s must be in location 12.) Either the p or the s can come first, so there are
2 * 7 = 14
ways to place the p and s.

Now, can you figure out how many ways you can arrange the remaining letters in "ermutation"?

(I'm assuming that by "in such a way that there always occur 4 letters between p and s" you mean that there are EXACTLY 4 letters between p and s, not at least 4.)

4. Hello, shivani!

How many ways we can arrange the word PERMUTATIONS
in such a way that there are always 4 letters between P and S?

We have 12 letters, including two T's.

The P and S can be placed: . $\begin{array}{c}P\:\_\:\_\:\_\:\_\:S \\ \text{or} \\S\:\_\:\_\:\_\:\_\:P\end{array}\quad\hdots$ . 2 ways

The pair can be placed at the left end ... to the right end: .7 ways.

The other 10 letters can be placed in: . ${\color{blue}{10\choose2}}$ ways.

Therefore, there are: . $2 \times 7 \times {10\choose2} \;=\;\boxed{25,401,600} \text{ arrangements.}$