Question 360323: If the letters of the word " REGULATIONS " be arranged at random, what is the chance that there will be exactly 4 letters between the R and the E
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the word is REGULATIONS
since there are 11 characters in the word, then those characters can be arranged in 11! = 39916800 ways randomly.
the number of ways that you can get 4 letters between any 2 specified letters are:
?xxxx?xxxxx
x?xxxx?xxxx
xx?xxxx?xxx
xxx?xxxx?xx
xxxx?xxxx?x
xxxxx?xxxx?
? represents the letters that have to have 4 letters between them.
x represents the other letters.
that's a total of 6 ways.
if you wanted to solve this part of problem using a formula, you would need to consider the number of ways as follows:
2 letters with 4 letters between them equals 1 block of 6 letters that have to be considered as 1 special letter because they are always together.
you start with 11 letters. you have 1 block of 6 letters that needs to be considered as 1.
subtract 6 from 11 and you get 5 remaining letters.
your letters are B x x x x x
The B represents the block of 6 letters.
The total possible ways this can happen is 6! / (1! * 5!) = 6! / 5! = 6.
Those possible ways are:
B x x x x
x B x x x
x x B x x
x x x B x
x x x x B
if you look at the 6 ways that I showed you and you replace ?xxxx? with B, then you will see that the B represents ?xxxx?.
So the number of ways that the pattern can present itself is 6.
the number of possible combinations in each one of those ways is:
9! * 2! = 725760.
the number of possible combinatons in all 6 of those ways is therefore:
6 * 9! * 2! = 6 * 725760 = 4354560
the probability that will happen is therefore equal to 4354560 / 39916800 = .109090909
hard to see with big numbers so let's try smaller numbers:
assume the word is MARK.
what is the probability that the A and the K will always have 1 letter between them.
total possible combinations in the word MARK equals 4! = 24.
they are:
MARK
MAKR
MRAK
MRKA
MKAR
MKRA
ARKM
ARMK
AKRM
AKMR
AMRK
AMKR
RMAK
RMKA
RAKM
RAMK
RKAM
RKMA
KMAR
KMRA
KAMR
KARM
KRMA
KRAM
now restrict the letters so there is exactly 1 letter betweeen A and K.
this can happen in 2 ways.
they are:
?x?x
x?x?
? represents either A or K
x represents either of the remaining letters.
total possible combinations are therefore equal to 2 * 2! * 2! = 8.
those possible combinations are:
A and K are in the 2d and 4th positions.,
MARK
MKRA
RAMK
RKMA
A and K are in the 1sd and 3d positions.
AMKR
KMAR
ARKM
KRAM
total number of possible ways is 4! = 24
total number of ways A and K can have exactly one letter between them is 2*2!*2! = 8.
probability of that happening is 8 / 24 = 1/3.
same concept only smaller numbers that can be shown easier.
`
|
|
|
