Permutations: 4 Balls in 7 Boxes with Maximum 1 Ball Each - Verify Answer

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Homework Statement



The number of ways in which 4 distinct balls can be kept in 7 different boxes if each box can have atmost 1 ball are?

The Attempt at a Solution



Easy one but I need to verify my answer.
The first ball can be kept in any of the 7 boxes in 7 ways
The second ball can be kept in 6 ways...
Permuting this way 7P4 = 840
Now, since the balls are distinct, total no. of ways are 840 x 4 = 3360

But the answer says it is 840.
Who is right?
 
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840 is correct. Suppose the balls are red, green, blue, and yellow. The red ball can be in anyone of 7 boxes, the green ball in anyone of the 6 remaining boxes, etc. So the answer is 7*6*5*4 = 840.

If the balls were not distinct, then you would divide 840 by 4! to get the answer in that case.
 
Avodyne said:
840 is correct. Suppose the balls are red, green, blue, and yellow. The red ball can be in anyone of 7 boxes, the green ball in anyone of the 6 remaining boxes, etc. So the answer is 7*6*5*4 = 840.

The first red ball which has a choice of 7 boxes, can be replaced by other colours too (green, blue or yellow).

i.e. Red ball can go to anyone of 7 boxes
Green...6 boxes
Blue...5 boxes
Yellow...4 boxes


Yellow ball can go to anyone of 7 boxes
Red...6 boxes
Green...5 boxes
Blue...4 boxes

...


There are 4 such cases. So we multiply the final result by 4.
Do you get my problem there?
 
Hi Abdul! :smile:
Abdul Quadeer said:
Red ball can go to anyone of 7 boxes
Green...6 boxes
Blue...5 boxes
Yellow...4 boxes

Yellow ball can go to anyone of 7 boxes
Red...6 boxes
Green...5 boxes
Blue...4 boxes

You're counting everything twice …

every arrangement is included in both your methods.
There are 4 such cases. So we multiply the final result by 4.

Why stop at 4? 4 is only the number of ways of choosing the first ball … why not multiply it by 3 and 2 also, for the second and third ball? :wink: (of course, that would count every arrangement 24 times!)
 
I understood it now. Thanks!