Six different balls are put in three different boxes, no box being emp

Question

Six different balls are put in three different boxes, no box being empty. What is he probability of putting balls in the boxes in equal numbers ?

(A) 1/28
(B) 1/21
(C) 1/8
(D) 1/7
(E) 1/6

Six different balls are put in three different boxes, no box being empty. The probability of putting balls in the boxes in equal numbers is

andreagonzalez2k · Accepted Answer

You have to count how balls are arranged inside boxes because the number is different depending of the balls that each box has.

If no box can be empty we have 3 different cases:

- 1 ball in one box, 2 balls in another box, 3 balls in another box
we can arrange boxes in 3! = 6 ways
and we can arrange balls inside boxes in 6C3 x 3C2 x 1C1 = 20 x 3 x 1 = 60
total = 6 x 60 = 360 ways

- 1 ball in one box, 1 ball in another box, 4 balls in another box
we can arrange boxes in 3!/2! = 3 ways
and we can arrange balls inside boxes in 6C4 x 2C1 x 1C1 = 15 x 2 x 1 = 30
total = 3 x 30 = 90 ways

- 2 balls in one box, 2 balls in another box, 2 balls in another box
we can arrange boxes in 3!/3! = 1 way
and we can arrange balls inside boxes in 6C2 x 4C2 x 2C2 = 15 x 6 x 1 = 90
total = 3 x 30 = 90 ways

Probability = 90/(360+90+90) = 90/540 = 1/6

Answer = E

sujoykrdatta · Answer

We are looking at distribution of 6 distinct objects in 3 distinct boxes such that no box is empty.

Each object has 3 options (boxes), hence total options = 3^6 = 729

However, this includes cases when boxes are empty. We need to remove those.

# Cases where 2 boxes are empty - there are only 3 cases (all 6 balls are in a single box) ... (i)

# Cases where 1 box is empty: the 2 boxes that are full can be chosen in 3C2 = 3 ways. 
Now, each ball has 2 options => number of cases = 2^6. 
However, this also includes cases when one box is empty; we need to remove that. This is possible in 2 ways only. 
Thus, cases = 2^6 - 2 = 62
Total cases (considering the fact that we need to select the 2 boxes in 3 ways) = 3*62 = 186 ... (ii)

Total cases = 729 - (3+186) = 540

Favorable cases = number of ways in which the 6 balls can be equally distributed in the 3 boxes; ie each box will have 2 balls: 
6C2 - choose 2 balls for the first box
4C2 - choose 2 balls for the second box
2C2 - choose 2 balls for the third box
=>  Number of favorable cases = 6C2 * 4C2 * 2C2 = 90

Thus, probability = 90/540 = 1/6

Answer E

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ZulfiquarA · Answer

Need some help

I got the answer 1/7
I want to know what I did wrong.
I made three scenarios of dividing 6 balls into
Scenario 1: 2*2*2 (Equal Digits) = 8
Scenario 2: 4*1*1 = Now this can can be rearranged in 3 ways so= 4*1*1*3=12
Scenario 3: 3*2*1 = This can be rearranged in 6 ways= 3*2*1*6=36

Probability=8/8+12+36 = 8/56=1/7  (HELP)

LaveenaPanchal · Answer

If no box be empty then there are 3 options

First = 4-1-1 = 6C4 x 2C1 x 1C1 = 15 x 2 = 30. This is possible in 3C2 x 1C1 ways i.e. 30 x 3 = 90 ways.
Second = 1-2-3 = 6C1 x 5C2 X 3C3 = 60. This is possible in 3!. 6 ways. i.e. 60 x 6 = 360 ways.
Third 2-2-2 = 6C2 x 4C2 x 2C2 ways. = 90 ways.

So the answer is 90/(90+180+90)=90/540 = 1/6

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LaveenaPanchal · Answer

..

I did it the same way but it took a lot of time to figure out..

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Fdambro294 · Answer

(Step 1) determine the different scenarios we can have in which the 6 balls can be placed in groupings such that no group will be empty (so no box will be empty)

(2nd) DEN = Total possible ways we can distribute the balls such that no box goes empty

Scenario 1:

(a) find the number of ways we can choose the 6 different balls such that they are placed into identical groupings of 4 - 1 - 1

“6 choose 4” = 6! / (4! 2!)
and
“2 choose 1” = 2! / (1! 1!)
and
“1 choose 1” = 1! / (1! 0!)

When you multiply these and cancel common Factorials in the NUM and DEN you will have:

6! / (4! 1! 1!)

However, since we have Two groups with identical sizes, when we choose we will be over-counting certain distributions. For instance, the grouping of:

——-   ——

and

——-  ——

Will each be counted when we do the choosing. However, for purposes of splitting be balls up into identical groups we do not care right now about the order.

Therefore, for every 2! Counts we have included, we only want to actually count 1 ——> thus, divide by (2!)

6! / (4! 1! 1!) * (1 / 2!) = “6 choose 4” = 15 ways

AND

(b) for each one of those ways, now that we have balls split into groups of 4 -1 - 1, we can arrange the 3 groupings among the 3 different boxes in ———> 3! Ways

(15) * 3! = 90 ways in total for scenario 1

Scenario 2:

Following the same logic as above:

(a) 6! / (3! 2! 1!) = 60 ways

AND

(b) 3! Ways to arrange among 3 different boxes

(60) * 3! = 360 ways for scenario 2

Scenario 3:   ***** the Favorable Outcome****

Because we have three groupings that will have identical sizes, for every 3! counted, we only way to keep 1.

However, once we put the balls into identical groupings of 2-2-2, we can then arrange the groupings among the 3 different boxes in 3! Ways.

3! Will cancel out.

(a) 6! / (2! 2! 2!) * (1 / 3!)

AND

(b) 3! ways to arrange

6! / (2! 2! 2!) * (1 / 3!) * (3!) =

6! / (2! 2! 2!) = 6 * 5 * 3 = 90 ways for scenario 3

Probability = (90) / (90 + 360 + 90) = 90/540 = 1/6

Answer: 1/6

(E)

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aliquidnihil · Answer

Interesting question about distributing 6 distinct balls into 3 distinct boxes with none empty. The key is to count all possible assignments (that meet the “no box empty” condition) and then count the favorable ones where each box has exactly 2 balls.

Kinshook · Answer

Six different balls are put in three different boxes, no box being empty.

What is the probability of putting balls in the boxes in equal numbers ?

Since, there is at least 1 ball in each box, let the extra balls in first, second and third boxes be x, y & z respectively.

Case 1: {1,1,4}
Possible arrangements = 6C1*5C1*4C4*3 = 90

Case 2: {1,2,3}
Possible arrangements = 6C1*5C2*3C3*3! = 6*10*6 = 360

Case 3: {2,2,2}
Possible arrangements = 6C2*4C2*2C2 = 15*6 = 90

Total arrangements = 90 + 360 + 90 = 540

Favorable arrangements with equal balls in each box = 6!/2*2*2 = 90

The probability of putting balls in the boxes in equal numbers = 90/540 = 1/6

IMO E

Six different balls are put in three different boxes, no box being emp

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