1) In how many ways can letters of the word $\text{MAMMAL}$ be arranged in a line?

1) In how many ways can letters of the word MAMMAL be arranged in a line?

For this question I put $$\frac{6!}{3!2!} = 60$$ since there are $3$ 'M's and $2$ 'A's

2) The letters of the word PROBABILITY are arranged at random. Find the probability that the two 'I' are separated.

Do I have to group the two I's to find the probability of them being separated ?

Thank you.

• You should fix the tag though, this ain't stats – Nikunj Feb 28 '16 at 21:44
• I do computer studies at College and Statistics is one of the courses I do. This is one of the subjects we do in stats. – Ken Feb 28 '16 at 21:46
• maybe but I thought that this was more of a combinatorics and probability problem – Nikunj Feb 28 '16 at 21:47
• Yes, that is a common misconception. Just because the class is called 'statistics' does not mean that the problem you are currently doing deals with actual statistics. Don't worry, it's not a real offense. Also, make sure to make your titles informative, like I did. Also also, the site uses MathJax formatting. Formatting tips here. – Em. Feb 28 '16 at 21:50
• $\text{Also}^3$, you're on the right track. For 2) It might be easier to calculate $1-P(\text{the two I's are together})$. – Em. Feb 28 '16 at 21:52

You can determine the number of distinguishable arrangements of the word PROBABILITY using similar reasoning.

$$\frac{11!}{2!2!}$$

From these arrangements, we must exclude those in which the two I's are consecutive. To do this, treat the two I's as a single letter. We then have ten objects to arrange, one of which is the double I.

$$\frac{10!}{2!}$$

To find the number of distinguishable arrangements of PROBABILITY in which the two I's are separated, subtract the number of arrangements in which they are together from the total number of arrangements.

$$\frac{11!}{2!2!} - \frac{10!}{2!}$$

• Could you tell how you hide the answer in the yellow background, I mean what do you type? It looks cool. – Nikunj Feb 28 '16 at 22:28
• @Nikunj You type >! before what you wish to hide. – N. F. Taussig Feb 28 '16 at 22:30

For 2), you can first count the number of ways to arrange the letters other than the I's,

$\hspace{.4 in}$ which is given by $\displaystyle\frac{9!}{2!}$ since there are 9 letters and two B's.

These 9 letters create 10 gaps, and there are $\dbinom{10}{2}$ ways to choose the two gaps for the I's.

Since there are $\displaystyle\frac{11!}{2!2!}$ ways to arrange the letters of PROBABILITY,

this gives a probability of $\displaystyle\frac{\frac{9!}{2!}\binom{10}{2}}{\frac{11!}{2!2!}}=\frac{\frac{9\cdot10!}{2!2!}}{\frac{11!}{2!2!}}=\frac{9}{11}$.