# Probability of a 12 symbol string containing 3 even digits

From Carol Ash's, "Probability Tutoring Book", pg. 19, prob. 1-3.4.

If a 12 symbol string is formed from the 10 digits and 26 letters, repetition not allowed, what is the prob that it contains 3 even digits?

I assumed string implies order matters, so I counted of $P(36,12)=\frac{36!}{(36-12)!}$ total possible strings.

Then for the numerator I counted:

1. Pick 3 spots for the 3 even digits
2. Fill the first, second and third even digit slots
3. From the remaining 31 non-even number characters pick something for each of the remaining 9 slots.

So I got $\frac{\binom{12}{3}\times P(5,3)\times P(31, 9)}{P(36,12)}$ where $P(n,k)=\frac{n!}{(n-k)!}$.

However Ash says:

For the total, pick a committee of 12 symbols from the 36. For the fav, pick a subcommittee of 3 evens and a subcommittee of 9 others. $$\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}$$

My question is:

1. Did I misunderstand the question and order doesn't matter?
2. Order does matter and I'm just wrong
3. These actually work out the same ? ( if so why? )
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They are the same but IMHO for such a solution the text should be "If a 12 symbol string is formed from the 10 digits and 26 letters, repetition not allowed, what is the prob that it contains exactly 3 even digits?". Otherwise I would assume an 'at least' is implied.Anyway I am not a native English speaker and the book surely knows better than me :) – AndreasT Dec 30 '12 at 10:20
@AndreasT It is somewhat ambiguous. But it's just a matter of getting used to the particular author's writing style. As long as they're consistent it's not too much of a problem. – Robert S. Barnes Dec 30 '12 at 10:43

They’re the same. Since $$P(n,k)=\binom{n}kk!\;,$$ we have

\begin{align*} \frac{\binom{12}{3}P(5,3)P(31, 9)}{P(36,12)}&=\frac{\binom{12}3\binom533!\binom{31}99!}{\binom{36}{12}12!}\\\\ &=\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}\cdot\frac{\binom{12}33!9!}{12!}\\\\ &=\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}\cdot\frac{12!}{12!}\\\\ &=\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}\;. \end{align*}

You’re quite right that order matters, but its effect on numerator and denominator is identical, so in the end it has no effect. Look at it this way: Ash’s computation has the number of acceptable sets of symbols in the numerator and the number of possible sets of symbols in the denominator. Each set of symbols can be permuted in $12!$ ways, so there are $$\binom53\binom{31}912!$$ acceptable strings of symbols and $$\binom{36}{12}12!$$ possible strings of symbols. That’s the ratio that you calculated. The numbers of acceptable and possible objects have changed, but the ratio itself has not.

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Here is one way of looking at it:

The total number of strings is $\frac{36!}{24!}$.

First pick 3 of 5 even digits. Order doesn't matter yet, so there are $\binom{5}{3}$ ways. Pick the remaining 9 digits from the remaining 31 non-even digits, order still doesn't matter, so there are $\binom{31}{9}$ ways.

Now place the 3 even digits. There are $(12)(11)(10) = \frac{12!}{9!}$ ways of doing this, and there are $9!$ ways of placing the remaining 9.

So the probability is $\frac{\binom{5}{3}\binom{31}{9}\frac{12!}{9!} 9!}{\frac{36!}{24!}} = \frac{\binom{5}{3}\binom{31}{9}}{\frac{36!}{12!24!}}= \frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}$.

One way of looking at it is that when you select, order does not matter, but when you finally place an object, order matters.

In your Step 1, you are selecting, so order does not matter. In Step 2, you are placing these objects, so order matters. In Step 3, you are combining selecting and placing in one step (select 9 out of 31 objects and place them into 9 slots). The combining cancels relevant quantities (and possibly confuses?).

I separated the selection from the placing.

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I really like the separation of selecting from placing. It makes the overall process much clearer. – Robert S. Barnes Dec 30 '12 at 10:39