Combinations and Permutations: 3 letters, repeat exactly three times

Question

If I have three letters, A, B, and C, and I want to use each letter exactly three times (9 places), what is the probability that I randomly pull out a nine-string of letters that starts with ABC?

I need help figuring out

How to find out how many possibilities there are for nine places, three letters that can be repeated exactly three times
What is the probability that my randomly selected string will start with ABC.

advocateofnone · Answer 1 · 2015-02-16T18:10:21.880

1

Total number of possibilities is $T=\frac{9!}{3!3!3!}$. Number of cases where string starts with "ABC" is $S=\frac{6!}{2!2!2!}$. So probability is $\frac{S}{T}$. All this comes from the basic formula that if we have $n$ items with $r_1$ of one type, $r_2$ of another and so on such that $r_1+r_2+...r_k=n$. Then number of distinct permutation of these $n$ items is given by $\frac{n!}{r_1!r_2!...r_k!}$.

edited Feb 16 '15 at 18:10

answered Feb 16 '15 at 18:05

advocateofnone

2,059

How did you know that it was 9!/ 3!3!3! and likewise 6!/2!2!2! – danielle Feb 16 '15 at 18:07
No, I'm just wondering if you could explain how you got to the formula for T and S. – danielle Feb 16 '15 at 18:09
See I mentioned a basic formula of combinations , permutation of similar objects, you can easily see the proof or google the proof – advocateofnone Feb 16 '15 at 18:11
Was I able to clear your doubt ? – advocateofnone Feb 16 '15 at 18:15

score 0 · Answer 2 · answered Mar 10 '15 at 01:44

There are $9!$ ways to arrange $9$ objects. If $3$ are the same, then $1/3!=1/6$ of these will be the same. For example consider $\{a_0,a_1,b\}, \{a_1,a_0,b\}, \{a_0,b,a_1\}, \{a_1,b,a_0\}, \{b,a_0,a_1\}, \{b,a_1,a_0\}$. There are $3!=6$ possibilities, but if we let $a_1=a_0$, then we only have $3$ possibilities, which is $3!/2!$. So with $\{A,A,A,B,B,B,C,C,C\}$ there are $9!/(3!3!3!)=1680$ possibilities.

Of these, $3/9=1/3$ start with an $A$, of these $3/8$ have a $B$ second, and $3/7$ have a $C$ third. So the final answer is $9!/(3!3!3!) \times (3.3.3)/(9.8.7)=6!/(2!2!2!)=90$.

Combinations and Permutations: 3 letters, repeat exactly three times

2 Answers2