# Simple question relating to probability

• kevinf
In summary, the problem is asking for the number of arrangements of "TOYBOAT" without the T's being next to each other. The first step is to find the total number of arrangements without the restriction, which is 7!/(2!*2!) due to the repeated letters of T and B. To ensure that the T's are not next to each other, they can be treated as one letter, resulting in 7 possible positions for the "T" letter. This would give a total of 7 x 5! arrangements, taking into account the repeated letter "O."

#### kevinf

Hi, i have a problem that asks how many arrangements of "TOYBOAT" are there if the T's can not be next to each other.
I know the first step is to find the total without the restriction, which is 7!/(2!*2!). the 2 2! represents the repeated letters of T and B but I'm not sure how to make it so that the T's can not be next to each other. I've listed the different ways that the T's could sit so that they are not next to each other, which is 30. any hints guys? it seems simple but for some reason i can't wrap my head around it

hi kevinf!

it often helps to go for the opposite

in this case, to find the number of ways in which the Ts are next to each other!

sorry but I'm not quite understsanding how i would get the answer that way but it would be 6 ways? and each of the 6 ways have 5! ways of arranging the other letters? or maybe not, since the O's are also repeated.

sorry lol can you elaborate on that a little more? sorry

yup … treat the two Ts as one letter

So then it would be 7 possible places that the t could sit in then. Then wouldn't it be 7 x 5! . But what about the repeated a.

kevinf said:
So then it would be 7 possible places that the t could sit in then.

uhh?

think again!

Lol if I understood you correctly, after making t one letter wouldn't there be 7 spots where t could go instead of 6 because t is now one letter

TT O Y B O A … only 6 letters!