Happy Easter!
If we randomly scramble the letters of HAPPYEASTER, what is the the probability the scramble will result in the A next to the A, the E next to the E, and the P next to the P?
Example of Desired Outcome: AAHYSPPTEER
Let the probability be: P = b a
where a and b are positive coprime integers.
Enter the answer as a + b
The answer is 499.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Hi David! Thanks for the solution. Do you know how to get the number of ways where exactly 2 of the doubled sets are next to each other ( Ex: AAPHYPEESTR ). I tried to write a problem, but then I realized my attempt was quite naïve.
Log in to reply
I believe you could do 3 for the number of ways to choose one of the three doubled letters to not be beside each other, then 2 ! 9 ! for the number of arrangements for H A 2 P P Y E 2 S T R , but then subtract out 8 ! for counting the number of rearrangements of H A 2 P 2 Y E 2 S T R where the two doubled sets that shouldn't be beside each other are beside each other.
So that would be 3 ( 2 ! 9 ! − 8 ! ) = 4 2 3 3 6 0 ways.
With 1 1 letters total and 3 letters that are repeated 2 times, the letters of H A P P Y E A S T E R can be rearranged in 2 ! 2 ! 2 ! 1 1 ! ways.
Treating double letters as a single letter, H A 2 P 2 Y E 2 S T R has 8 different letters and can be rearranged in 8 ! different ways.
The probability is then 2 ! 2 ! 2 ! 1 1 ! 8 ! = 1 1 ⋅ 1 0 ⋅ 9 8 = 4 9 5 4 , so a = 4 , b = 4 9 5 , and a + b = 4 9 9 .