Answer
a. 151,200 ways b. 10,080 ways c. 10,080 ways
Explanation:
a. How many different strings can be made from the letters in the work PEPPERCORN when all the letters are used?
Since there are 10 letters in the word PEPPERCORN, there are 10! ways of arranging them. Since we have 3 P's, there are 3! ways of arranging them . We have 2 E's, there are 2! ways of arranging them. We have 2 R's, there are 2! ways of arranging them. We have 1 O, there are ! ways of arranging them. We have 1 C, there are 1! ways of arranging them. We have 1 N, there are 1! ways of arranging them. So there are
[tex]\frac{10!}{3!2!2!1!1!1!} = 151,200 ways[/tex] of arranging the word
b. How may of the strings start and end with the letter P?
If the strings start and end with P, then we have 8 letters left including 1 P.
Since there are 8 letters left in the word PEPPERCORN, there are 8! ways of arranging them. Since we have 1 P, there are 1! ways of arranging them . We have 2 E's, there are 2! ways of arranging them. We have 2 R's, there are 2! ways of arranging them. We have 1 O, there are ! ways of arranging them. We have 1 C, there are 1! ways of arranging them. We have 1 N, there are 1! ways of arranging them. So there are
[tex]\frac{8!}{2!2!1!1!1!1!} = 10,080 ways[/tex] of arranging the word
c. How many strings have 3 consecutive Ps?
There are 8 different ways of arranging the P's consecutively. We are then left with 7 letters. Since there are 7 letters left in the word PEPPERCORN, there are 7! ways of arranging them. We have 2 E's, there are 2! ways of arranging them. We have 2 R's, there are 2! ways of arranging them. We have 1 O, there are ! ways of arranging them. We have 1 C, there are 1! ways of arranging them. We have 1 N, there are 1! ways of arranging them. So there are
[tex]\frac{8!}{2!2!1!1!1!1!} = 10,080 ways[/tex] of arranging the word
