Bunuel wrote:
Marshall has 4 different colored balls and 3 different boxes. In how many ways can he put these balls in the boxes so that no box is left empty and no ball is left out?
A. 18
B. 24
C. 36
D. 72
E. 81
Are You Up For the Challenge: 700 Level Questions4 balls: A, B, C, D
3 distinct boxes: we can have the 1st box = 1st place; 2nd box = 2nd place; and 3rd box = 3rd place
If we were to first choose 3 balls out of 4 ——-then for each distinct group of 3 balls arrange them in 3! ways throughout the 3 places ——and then finally place the remaining ball in 1 of 3 boxes
We would end up with:
(4 c 3) * (3!) * (3) = 4 * 6 * 3 = 72 ways
To show why we would be double counting each case we can run through two possibilities:
Case 1: for the step 1, we end up choosing a group of 3 balls that is: (A—B—C) and 1 left over is (D)
There would be 3! = 6 different Ways to arrange them in the 3 places/boxes
A-B-C
A-C-B
etc.
the one left over ball is D - and for each one of these 6 ways, we would place the ball in 3 different places/boxes.
Let’s just look at the first way: (A-B-C)
We would have:
(A-D) — (B) — (C)
(A) — (B-D) —(C)
(A) — (B) — (CD)******
Case 2: this time, instead, for step 1 we end up choosing a group of 3 balls that is: (A—B—D) and 1 left over is C
Again there would be 3! = 6 different ways to arrange this group in the 3 places/boxes
A-B-D
A-D-B
etc.
The one left over is C — and we could arrange this in 3 ways for each one of the above 6 permutations. Let’s just choose 1 of the 6: (A—B—D)
We would have:
(A-C) — (B) — (D)
(A) — (B - C) — (D)
(A) — (B) — (D-C)****
However, look at the 2 starred ways:
The only difference is the ORDER in which we placed Balls D and C in the 3rd box. And since the order of the balls in each box does NOT MATTER, we have double counted the number of ways the distribution of balls to boxes can be done.
72 would be the correct answer if the ORDER MATTERED of the 2 balls that are placed together in the box
so you can “unorder” the 72 arrangements by dividing the 72 ways by 2! (Since for every 1 unique arrangement we want to keep we accidentally counted 2). This will get you to the correct answer of 36
or
From the beginning, in order to avoid accidentally ordering/arranging the 2 balls that end up in the box together, you can follow the below procedure.
Step 1: out of the 4 balls, first find how many ways we can choose an unordered group of 2 balls to be placed in a box together.
(4 c 2) = 6 ways
AND
Step 2; for each of these 6 ways, there are 3 choices of boxes in which you can place the unordered group of 2 balls:
(3 c 1) = 3 ways
AND
Step 3: finally, for the last 2 remaining balls, we can arrange these 2 balls in 2 boxes in:
2! = 2 ways
(4 c 2) * (3 c 1) * (2!) =
6 * 3 * 2 =
36 ways
C is the answer
Edit:
The answer above mine is also a clever way to get to the answer.
Step 1: You can figure out how many groups of 2 balls you can create out of 4. When you choose the 2 balls you want to include, you have automatically separated the other 2 balls from the group you’ve chosen.
(4 c 2) = 6 ways
for each of these 6 ways, we can assume the 2 unchosen balls are each part of their own separate group—— so that we would have 3 groupings of balls for each one of the 6 cases
These 6 cases would be grouped as follows:
(AB) - (C) - (D)
(AC) - (B) - (D)
(AD) - (B) - (C)
(A) - (BC) - (D)
(A) - (BD) - (C)
(A) - (B) - (CD)
AND
Step 2: for each of the 6 cases above, we have 3 boxes in which to arrange the 3 “distinct items”:
3! = 6 arrangements
(4 c 2) * (3!) =
(6) * (6) = 36 ways
Posted from my mobile device