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31 May 2020, 03:01
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
52% (01:45) correct
48%
(02:21)
wrong
based on 153
sessions
History
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In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each?
A.84
B.70
C.120
D.56
E.54
A.84
B.70
C.120
D.56
E.54
Updated on: 22 Feb 2025, 00:14
Originally posted by GMATinsight on 31 May 2020, 04:52.
Last edited by Bunuel on 22 Feb 2025, 00:14, edited 2 times in total.
Last edited by Bunuel on 22 Feb 2025, 00:14, edited 2 times in total.
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MBADream786
In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each?
A.84
B.70
C.120
D.56
E.54
A.84
B.70
C.120
D.56
E.54
\(P+Q+R+S = 7\)
\(P_{min} = 1\)
\(Q_{min} = 1\)
Let's give one ball in each box P and Q, so now we have 5 more balls to distribute among 4 Boxes P, Q, R and S
i.e. New Equation, P+Q+R+S = 5
Now, we have find non-negative solution to this equation
For any equation a+b+c+d = n
where n = Total Balls to be distributed and r = 4 (Number of variables)
The number of Non-negative solutions \(= (n+r-1)C_{r-1}\)
i.e. Total Non-negative Integer solutions of P+Q+R+S = 5 will be \((5+4-1)C_{4-1} = 8C2 = 56\)
Answer: Option D
DERIVATION OF PARTITION RULE:
31 May 2020, 03:45
Kudos
Bookmarks
MBADream786
In how many ways can 7 identical balls be placed into four boxes P,Q,R,S such that the two boxes P and Q have atleast one ball each?
A.84
B.70
C.120
D.56
E.54
A.84
B.70
C.120
D.56
E.54
As P and Q have to have at least one ball, let us give one each to P and Q. Now we are left with 5 balls and they have to be distributed amongst P, Q, R and S.
So P+Q+R+S=5
Now the direct formula of putting three partitions in 5 balls makes it 5+3 or 8 out of which we have to select these 3 partitions. => 8C3=8*7*6/3!=56
D
Since 5 is not a big number, we can find each way of distribution as
1) 5,0,0,0 —4 way
2) 4,1,0,0 — 4!/2!=12
3) 3,2,0,0 — 12 ways
4) 3,1,1,0 — 12 ways
5) 2,2,1,0– 12 ways
6) 2,1,1,1– 4 ways
Total 56 ways
