If the letters TRAPDOOR are randomly arranged in a row, find the probability that the two R's are adjacent and the two O's are adjacent.

Guest Jun 24, 2020

#1**+1 **

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See TLB answer below !

Thanx , TLB for the clarification below !!!!

ElectricPavlov Jun 24, 2020

#2**0 **

The total number of ways to arrange TRAPDOOR is \(\frac{8!}{2!*2!}\). 8 choices for the first letter, 7 choices for the second etc. We divide by 2!*2! because there are repeating cases with the two R's and the two O's.

Now onto the cases that the two R's are adjacent and the two O's are adjacent.

Think of the two R's as a "block" or a "bundle". Think of the two O's the same way.

So now we find the number of ways to arrange 6 letters (The 4 distinct letters T, A, P, D & the 2 "bundles" of R and O). This is just 6!. We don't have to divide by anything because they are all distinct.

\(\frac{6!}{\frac{8!}{2!*2!}}\)= \(\frac{1}{14}\)

thelizzybeth Jun 24, 2020

#3**0 **

Thanx, TLB but I think you calc'd oo and rr being ANYWHERE in the sequence....not the probability of them being NEXT to each other??

ElectricPavlov
Jun 24, 2020

#4**0 **

The question asks __that the two R's are adjacent__ **and** __the two O's are adjacent__. RR and OO don't have to be adjacent to each other. It could be xooxxrrx

I think you intepreted it as "The two R's and the two O's are adjacent". Then you would be correct.

thelizzybeth
Jun 24, 2020

#7**0 **

TRAPDOOR =8 letters =8! =40,320 permutations. But, we have 2 repeated pairs [O,O and R,R], so we have a total of 8!/2!.2! =10,080 permutations. Each permutation begins with a letter, A,D,O,O,P,R,R,T. Each permutation that begins with any letter continues with that letter for: 10,080 / 8 =1,260 permutations. And since you have 4 letters, out of total 8 letters, repeated, the two "Os" and the two "Rs" are going to be adjacent to each other:1,260 x 4 =5040 times out of a total of 10,080 permutations, or 5040 /10,080 =1/2.This makes sense since 1/2 of the 8 letters are repeated.

Guest Jun 25, 2020