Seven Boxes

There are 7 boxes each written with letter $A$ to $G$ ($A$ for the first box, $B$ for the 2nd and so on). You have 7 balls each with letter $A$ to $G$ ($A$ for the first ball, $B$ for the 2nd and so on). You randomly put one ball in each box (blindfolded). Find the probability that you put 3 balls correctly (that is, you put the ball in the box with the same letter)?
Source: submitted by Jimel Mariano 9/25/2011

SOLUTION
We randomly put a ball in a box. We have 7 choices for the first box, then 6 for the second box, etc. The number of possible outcomes is $7\times 6\times 5\times 4\times 3\times 2\times 1=5040$.

Putting 3 balls correctly
We want to investigate the desirable outcomes of putting 3 balls correctly. Without loss of generality assume that we put 3 balls $A,B,C$ correctly in the 3 boxes labeled $A,B,C$ as shown in the diagram below
$\underline A\;\underline B\;\underline C\;\underline{\;\;\;}\;\underline{\;\;\;}\;\underline{\;\;\;}\;\underline{\;\;\;}$

The remaining balls are $D,E,F,G$. We cannot put ball $D$ in the fourth box because that would make 4 balls placed correctly instead of 3. But we can put one of the balls $E,F,G$ in the fourth box and ball $D$ somewhere else. We now consider the 3 cases depending on whether we place ball $E,F$ or $G$ in the fourth box.

Case 1: $\underline A\;\underline B\;\underline C\;\underline E\;\underline{\;\;\;}\;\underline{\;\;\;}\;\underline{\;\;\;}$
Since ball $E$ is already in the wrong place, we can put any of the remaining balls $D,F,G$ in the fifth box. We have 3 choices
$\underline A\;\underline B\;\underline C\;\underline E\;\underline D\;\underline G\;\underline F$        $F$ cannot be in sixth and $G$ cannot be in seventh
$\underline A\;\underline B\;\underline C\;\underline E\;\underline F\;\underline G\;\underline D$        $G$ cannot be in seventh
$\underline A\;\underline B\;\underline C\;\underline E\;\underline G\;\underline D\;\underline F$        $F$ cannot be in sixth

Case 2: $\underline A\;\underline B\;\underline C\;\underline F\;\underline{\;\;\;}\;\underline{\;\;\;}\;\underline{\;\;\;}$
As long as we don’t put ball $E$ in the fifth box we will be OK. We have 3 choices
$\underline A\;\underline B\;\underline C\;\underline F\;\underline D\;\underline G\;\underline E$        $G$ cannot be in seventh
$\underline A\;\underline B\;\underline C\;\underline F\;\underline G\;\underline D\;\underline E$        no restriction on $D,E$
$\underline A\;\underline B\;\underline C\;\underline F\;\underline G\;\underline E\;\underline D$

Case 3: $\underline A\;\underline B\;\underline C\;\underline G\;\underline{\;\;\;}\;\underline{\;\;\;}\;\underline{\;\;\;}$
As long as we don’t put ball $E$ in the fifth box we will be OK. We have 3 choices

$\underline A\;\underline B\;\underline C\;\underline G\;\underline {D}\;\underline E\;\underline F$        $F$ cannot be in sixth
$\underline A\;\underline B\;\underline C\;\underline G\;\underline F\;\underline D\;\underline E$        no restriction on $D,E$
$\underline A\;\underline B\;\underline C\;\underline G\;\underline F\;\underline E\;\underline D$

The number of desirable outcomes for putting 3 balls $ABC$ correctly equals 9.

Probability of putting 3 balls correctly
The same reasoning applies to any other three boxes. For example, the number of desirable outcomes for putting balls $B,D,F$ correctly in boxes labeled $B, D,F$ equals 9
. How many ways can we choose 3 boxes from a set of seven? The answer is given by the “choose number”
$_7C_3=\frac{7!}{3!\left (7-3\right )!}$
$=\frac{7!}{3!4!}$

$=\frac{7\cdot 6\cdot 5\cdot 4!}{3!4!}$
$=7\cdot 5$
$=35$

Therefore the number of desirable outcomes for putting ANY 3 balls correctly equals
$35\times 9=315$

The probability of putting 3 balls correctly is
$\frac{315}{5040}=\frac{1}{16}$

Answer: $\frac{1}{16}$