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03 Apr 2020, 05:09
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D
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Difficulty:
55%
(hard)
Question Stats:
63% (01:50) correct
37%
(01:47)
wrong
based on 206
sessions
History
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Not Attempted Yet
In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
A. 3
B. 4
C. 5
D. 6
E. 7
Are You Up For the Challenge: 700 Level Questions
A. 3
B. 4
C. 5
D. 6
E. 7
Are You Up For the Challenge: 700 Level Questions
03 Apr 2020, 06:50
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A gets at least 1 pen, B gets at least 2 pens and K gets at least 3 pens
So 1+2+3 = 6 pens are already lets say must be distributed in a certain way as specified
now remaining number of pens = 8-2 = 2 pens
These 2 pens can be divided among A,B and K in the following ways -
A B K
2 0 0
0 2 0
0 0 2
1 1 0
1 0 1
0 1 1
Thats it 6 ways. So answer is D
So 1+2+3 = 6 pens are already lets say must be distributed in a certain way as specified
now remaining number of pens = 8-2 = 2 pens
These 2 pens can be divided among A,B and K in the following ways -
A B K
2 0 0
0 2 0
0 0 2
1 1 0
1 0 1
0 1 1
Thats it 6 ways. So answer is D
General Discussion
03 Apr 2020, 06:09
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Bunuel
Formula: For all non-negative integers \(x_1\), \(x_2\), \(x_3\), . . . . . .\(x_r\), Number of solutions of \(x_1\) + \(x_2\) + \(x_3\) + . . . . \(x_r\) = n is \((n + r - 1)c_{r - 1}\)
Let the number of pens Amal, Bimal & Kamal gets = {\(a + 1, b + 2, k + 3\)}, for all non-negative integers \(a\), \(b\) & \(k\)
--> \(a + 1 + b + 2 + k + 3 = 8\)
--> \(a + b + k = 2\)
--> Number of distributions possible = \((2 + 3 - 1)c_{3 - 1} = 4c_2 = 6\)
Option D
