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parveen110
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Case 1: 1 box gets 3 balls, the other 4 boxes each get 1 ballparveen110 wrote:There are 5 different boxes and 7 different balls.All the balls are to be distributed in the 5 boxes placed in a row so that any box can recieve any number of balls. In how many ways can this ball be distributed so that no box is empty?
a. 7!
b. 16800
c. 1775
d. none of these
OA: B
Number of box options for the box with 3 balls = 5. (Any of the 5 boxes.)
From the 7 balls, the number of ways to choose 3 balls for this box = 7C3 = (7*6*5)/(3*2*1) = 35.
Number of ways to arrange the remaining 4 balls = 4! = 24.
To combine these options, we multiply:
5*35*24 = 4200.
Case 2: 2 boxes each get 2 balls, the other 3 boxes each get 1 ball
From the 5 boxes, the number of ways to choose a pair of boxes to hold 2 balls each = 7C2 = (5*4)/(2*1) = 10.
From the 7 balls, the number of ways to choose a pair of balls for the 1st box = (7*6)/*2*1) = 21.
From the remaining 5 balls, the number of ways to choose a pair of balls for the 2nd box = (5*4)/(2*1) = 10.
Number of ways to arrange the remaining 3 balls = 3! = 6.
To combine these options, we multiply:
10*21*10*6 = 12600.
Total ways = 4200+12600 = 16800.
The correct answer is B.
