How many ways can the letters of the word HAPPINESS be arranged. If H must start the word, N must end the word and the P's must be separated? [question from the topic FACTORIAL,PERMUTATION AND COMBINATION ]
Dennis was not a subject taught when I was in school.
90,720
90,720 9! / 2! 2! = 362,880/4 = 90,720
Well, let's break this down, one letter at a time! We have the word HAPPINESS with 9 letters. Since H must start the word and N must end it, we can treat them as fixed positions.
Now, we have 7 remaining letters to arrange: A, P, P, I, N, E, S. Since the two P's must be separated, we can think of them as distinct objects.
Starting with the 7 remaining letters, we have 7 options for the first position, 6 options for the second position, and so on. This gives us 7! (7 factorial) possibilities.
However, since we have two identical P's, we need to divide by 2! (2 factorial) to account for the overcounting.
So, the total number of ways we can arrange the letters of the word HAPPINESS with the given conditions is 7! / 2! = 5,040 ways.
That's quite a lot of HAPPINESS, don't you think?
To answer this question, we need to break it down into smaller parts.
First, let's consider that "H" must start the word and "N" must end the word. This means that we have fixed positions for these two letters. Therefore, we can ignore them for now and focus on the remaining letters: A, P, P, I, E, S, and S.
So, we need to arrange these 6 letters - A, P, P, I, E, S, and S - in such a way that the P's are separated.
To do this, we can start by arranging these 6 letters without any restrictions. This can be done using the concept of permutations. The formula for finding the number of permutations is given by nPn = n!, where "nPn" denotes the permutation of "n" objects taken "n" at a time, and "n!" represents the factorial of "n".
In this case, we have 6 distinct letters to arrange, so the total number of ways to arrange them without any restrictions is 6P6 = 6!.
However, since we have two identical P's, we need to correct for the overcounting. The two P's can be interchanged, but these arrangements would be considered the same. To find the total number of ways to arrange the 6 letters with the two P's, we divide by 2!.
Therefore, the number of ways to arrange the letters A, P, P, I, E, S, and S with the given condition is:
6! / (2! * 2!)
Simplifying this expression, we get:
(6 * 5 * 4 * 3 * 2 * 1) / (2 * 1 * 2 * 1)
= 720 / 4
= 180
So, there are 180 different ways to arrange the letters of the word HAPPINESS, with the conditions that H must start the word, N must end the word, and the P's must be separated.