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Question
Fifteen identical balls have to be put in five different boxes. Each box can contain any number of balls. The total number of ways of putting the balls into the boxes so that each box contains at least two balls is equal to k. Then which of the following is/are divisor of k:
Fifteen identical balls have to be put in five different boxes. Each box can contain any number of balls. The total number of ways of putting the balls into the boxes so that each box contains at least two balls is equal to k. Then which of the following is/are divisor of k:
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Solution
The correct options are
A 9
B 7
D 21
Let the balls put in the boxes named as
x1,x2,x3,x4 and x5. Then we have :
x1+x2+x3+x4+x5=15, xi≥2
⇒(x1−2)+(x2−2)+(x3−2)+(x4−2)+(x5−2)=5
⇒y1+y2+y3+y4+y5=5, yi=xi−2≥0;The total number of ways is equal to number of non-negative integral solutions of the last equation, which is = 5+5−1C5−1= 9C4=126
A 9
B 7
D 21
Let the balls put in the boxes named as
x1,x2,x3,x4 and x5. Then we have :
x1+x2+x3+x4+x5=15, xi≥2
⇒(x1−2)+(x2−2)+(x3−2)+(x4−2)+(x5−2)=5
⇒y1+y2+y3+y4+y5=5, yi=xi−2≥0;The total number of ways is equal to number of non-negative integral solutions of the last equation, which is = 5+5−1C5−1= 9C4=126
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