Question

How many different arrangements can be formed out of the letters of PRODUCTS, keeping P as the
first and S as the last letter?

Answer 1

Answer:

6! = 720

Step-by-step explanation:

since P and S are fixed only the order od the remaining 6 letters can be changed.

using permutations it can be written as

P(6,6) = 6!/1

= 720

Answer 2

Step-by-step explanation:

As per the data given in  the question,

We have to determine the number of arrangments by which the letters of word PRODUCTS can be rearranged keeping P and S constant.

As here,

There are total 8 letters in the given word.

So, if we put P at first place and S at last place, then we have only 6 positions left at which the 6 letters R, O, D, U, C, T are need to placed.

As we know,

n letters can be placed at n places by $n!$ ways.

So, 6 letters can be placed at 6 different places by $6!$ ways.

So, $6! = 6 \times 5\times 4\times 3\times 2\times 1=720$ ways.

Answer: Hence, 720 different arrangements can be formed out of the letters of PRODUCTS, keeping P as the first and S as the last letter.