Hello mormor83

Welcome to Math Help Forum! Originally Posted by

**mormor83** hello everyone

i am having problem with 2 question :

1) say i don't know the letters a-z and i need to create the word

" Probability" from the letters "P-r-o-b-a-b-i-l-i-t-y " . ( i have exactly 11 places )

2) i have 4 kids with different names but i don't know the names of either of them. what is the prob that i wont call either of them in the right name?

please help me(this forum rockz

)

mormor83

1) I think the question means: The letters "p-r-o-b-a-b-i-l-i-t-y" are arranged in a random order. What is the probability that they spell the word "probability"?

If this is what the question means, then you need to know the number of permutations (arrangements) of $\displaystyle 11$ items that contain $\displaystyle 2$ items repeated of the first kind ('b') and $\displaystyle 2$ items repeated of the second kind ('i'). This is$\displaystyle \frac{11!}{2!2!}$

(For a general formula see this page.)

Since just one of these arrangements is the correct one, the probability that this arrangement occurs at random is$\displaystyle \frac{2!2!}{11!}$

2) Again, if I may re-word the question: the names of 4 children are randomly selected by each child. What is the probability that none of the children selects their own name?

Such a selection - where no item is chosen in its 'natural' place - is called a *derangement*. With $\displaystyle 4$ items, there are $\displaystyle 9$ derangements. (See, for example, just here.)

It doesn't take long to list all the possibilities with 4 items, but if you want the formula for $\displaystyle d_n$, the number of derangements of $\displaystyle n$ items, it is in the form of a recurrence relation:$\displaystyle d_n=nd_{n-1}+(-1)^n, n\ge 1,$ with $\displaystyle d_0$ defined as $\displaystyle d_0=1$

You'll see that this gives:$\displaystyle d_1=0$

$\displaystyle d_2=1$

$\displaystyle d_3=2$

$\displaystyle d_4=9$

$\displaystyle d_5=44$

and so on.

Since the number of arrangements of $\displaystyle 4$ items is $\displaystyle 4!\,(=24)$, the probability that one chosen at random is a derangement is clearly $\displaystyle \frac{9}{24}=\frac38$

Grandad