P & C problem

parveen110 wrote:There are four different coloured balls and 4 boxes of the same colours as that of the balls. Find the number of ways in which exactly one ball can be put in a different colour box than that of the ball?
a. 16
b. 10
c. 4C2
d. none of these

Ans: d

The problem does not constrain how many balls can be placed in each box, implying that each box can hold more than one ball.

Let's say that the correct distribution of the 4 balls is as follows:
A-B-C-D.
In the correct distribution, A is in the 1st box, B is in the 2nd box, C is in the 3rd box, and D is in the 4th box.

Ways to put ONLY A in the wrong box:
empty-AB-C-D
empty-B-AC-D
empty-B-C-AD
Total ways = 3.

Extending this reasoning to the other 3 balls, there will be 3 ways to put ONLY B in the wrong box, 3 ways to put ONLY C in the wrong box, and 3 ways to put ONLY D in the wrong box.

Since for each of the 4 balls there are 3 ways to put exactly ONE ball in the wrong box, the total number of ways = 4*3 = 12.

The correct answer is D.

GMATGuruNY wrote:

parveen110 wrote:There are four different coloured balls and 4 boxes of the same colours as that of the balls. Find the number of ways in which exactly one ball can be put in a different colour box than that of the ball?
a. 16
b. 10
c. 4C2
d. none of these

Ans: d

The problem does not constrain how many balls can be placed in each box, implying that each box can hold more than one ball.

Let's say that the correct distribution of the 4 balls is as follows:
A-B-C-D.
In the correct distribution, A is in the 1st box, B is in the 2nd box, C is in the 3rd box, and D is in the 4th box.

Ways to put ONLY A in the wrong box:
empty-AB-C-D
empty-B-AC-D
empty-B-C-AD
Total ways = 3.

Extending this reasoning to the other 3 balls, there will be 3 ways to put ONLY B in the wrong box, 3 ways to put ONLY C in the wrong box, and 3 ways to put ONLY D in the wrong box.

Since for each of the 4 balls there are 3 ways to put exactly ONE ball in the wrong box, the total number of ways = 4*3 = 12.

The correct answer is D.

Thank you for the reply, Mitch. Solution is as usual lucid. However, I have a question. What if there was a constraint that each box can hold only one ball? Then I may figure out that not one, but two combinations will be different as one colour combination can never be wrong. Could you please extend this logic?
Thanks.

parveen110 wrote:

GMATGuruNY wrote:

parveen110 wrote:There are four different coloured balls and 4 boxes of the same colours as that of the balls. Find the number of ways in which exactly one ball can be put in a different colour box than that of the ball?
a. 16
b. 10
c. 4C2
d. none of these

Ans: d

The problem does not constrain how many balls can be placed in each box, implying that each box can hold more than one ball.

Let's say that the correct distribution of the 4 balls is as follows:
A-B-C-D.
In the correct distribution, A is in the 1st box, B is in the 2nd box, C is in the 3rd box, and D is in the 4th box.

Ways to put ONLY A in the wrong box:
empty-AB-C-D
empty-B-AC-D
empty-B-C-AD
Total ways = 3.

Extending this reasoning to the other 3 balls, there will be 3 ways to put ONLY B in the wrong box, 3 ways to put ONLY C in the wrong box, and 3 ways to put ONLY D in the wrong box.

Since for each of the 4 balls there are 3 ways to put exactly ONE ball in the wrong box, the total number of ways = 4*3 = 12.

The correct answer is D.

Thank you for the reply, Mitch. Solution is as usual lucid. However, I have a question. What if there was a constraint that each box can hold only one ball? Then I may figure out that not one, but two combinations will be different as one colour combination can never be wrong. Could you please extend this logic?
Thanks.

Dear parveen110,

This problem as a whole doesn't reflect the real GMAT, as there are only 4 choices and the nCr notation has been used in the choices. But, it's a nice problem. Aside, if there was a constraint that each box can hold only one ball, then this would have been an impossible case in which exactly one ball can be put in a different colour box than that of the ball. Still, D would have been a half-heart choice, since it doesn't clearly define the real situation. Hence, it's good for the problem that there's no restriction for the number of balls that a box may carry.

ceilidh.erickson wrote:

sanju09 wrote: Dear parveen110,

This problem as a whole doesn't reflect the real GMAT, as there are only 4 choices and the nCr notation has been used in the choices. But, it's a nice problem. Aside, if there was a constraint that each box can hold only one ball, then this would have been an impossible case in which exactly one ball can be put in a different colour box than that of the ball. Still, D would have been a half-heart choice, since it doesn't clearly define the real situation. Hence, it's good for the problem that there's no restriction for the number of balls that a box may carry.

You're right; this problem certainly does not reflect the GMAT.
- the GMAT will always have 5 choices
- nCr notation will never be used on the GMAT
- the question is confusingly written. It is not clear what they mean by "same colours as the balls," and we don't know how many balls can go in each.
- the GMAT will use American standard spelling: "colors," not "colours."

I have to disagree with your statement that "it's a nice problem," though. It's a very bad idea to study with non-GMAT-like problems! Even if some of the math is similar, it won't help to prepare you for interpreting the language and developing strategies for the actual test.

It's not helpful to study from materials that do not reflect the actual test!

Thanks for the half agreement, anyway!