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}

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About mvtrinh

Retired high school math teacher.
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