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In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 02:01
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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
Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 02:45
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MBADream786 wrote:
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
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
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Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 02:57
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MBADream786 wrote:
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
Though i got it wrong, here is my solution P(1), Q(1), (R+S)[5]=1*1*6( because 5 balls can be distributed among R and S in (0,5),(1,4),(2,3)*(2) as they can either be in R or S=6 ways [P+Q](3) R+S(4)= one ball in P , two ball in Q or Reverse so 2 ways * for R and S ( 0,4),(1,3),(2,2) 5 ways as 2,2 is same for R or S=10 ways P+Q=4, R+S=3 Total 12 ways P+Q=(1,3),(2,2),(3,1) and R+S=(0,3),(1,2),(2,1),(3,0) P+Q=5, R+S=2 Total 12 ways P+Q=(1,4), (4,1),(2,3),(3,2) R+S= (0,2),(2,0),(1,1) P+Q=6, R+S=1 Total 10 ways P+Q= (1,5),(5,1),(3,3),(2,4),(4,2) R+S=(0,1),(1,0) P+Q=7 R+S=0 Total 6 ways P+Q= (1,6),(6,1),(2,5),(5,2),(3,4),(4,3) Total=6+10+12+12+10+6 56 D:)
Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 03:05
chetan2u wrote:
MBADream786 wrote:
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
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
Sir can you explain this highlighted part. Why we put 3 partition in 5. When there are 4 boxes.
Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 03:13
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yashikaaggarwal wrote:
chetan2u wrote:
MBADream786 wrote:
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
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
Sir can you explain this highlighted part. Why we put 3 partition in 5. When there are 4 boxes.
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Hi
When we put 3 partitions, we get items in 4 places. oolololo.....lines show partition , so 2,1,1,1 loollooo.....so 0,2,0,3 And so on
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Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 03:24
MBADream786 wrote:
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
given that out of 7 balls P,Q need to have atleast one ball each so we put 1 ball each in P,Q left with 5 balls which can be arranged in 4 boxes P,Q,R,S in (5+4-1) c ( 4-1) ways or say 8c3 ways OPTION D ; 56
Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 03:26
Ruchirkalra wrote:
chetan2u wrote:
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
Can you please explain this a bit. Why are you doing 5+3?
There are 5 balls left and we can put them in any way. So ooooo, what we do is we take 3 partitions so that we can place those partitions between these 5 balls so that these 5 balls are distributed in 4 parts. ooooolll, so total we have 5+3 places and the 3 partitions can take any of the 8 locations. It can be lllooooo....0,0,0,5 loololoo....0,2,1,2 ollloooo....1,0,0,4 and so on
This is the direct method for such linear equations where you have sum of variables on one side and a numerical value on the other side
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In how many ways can 7 identical balls be placed into four boxes P,Q,R
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Updated on: 31 May 2020, 05:05
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2
MBADream786 wrote:
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
\(P+Q+R+S = 7\)
\(P_{min} = 1\) \(Q_{min} = 1\)
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:
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Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 04:55
MBADream786 wrote:
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
P 4.......2........3........5........4
Q 1.......2.........2.......2........2
R 1.......2.........1.......0.........1
S 1.......1.........1.......0.........0
So \(\frac{4!}{3!} + \frac{4!}{3!}+\frac{4!}{2!} +\frac{4!}{2!} +\frac{4!}{1! }= 4+4+12+12+24 = 56\)
So D
I wonder if there is a shortcut solution to this question, cause it took me more than 4 min to figure out the answer
Re: In how many ways can 7 identical balls be placed into four boxes P,Q,R
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31 May 2020, 05:03
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dave13 wrote:
MBADream786 wrote:
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
P 4.......2........3........5........4
Q 1.......2.........2.......2........2
R 1.......2.........1.......0.........1
S 1.......1.........1.......0.........0
So \(\frac{4!}{3!} + \frac{4!}{3!}+\frac{4!}{2!} +\frac{4!}{2!} +\frac{4!}{1! }= 4+4+12+12+24 = 56\)
So D
I wonder if there is a shortcut solution to this question, cause it took me more than 4 min to figure out the answer
This is a typical question based on partition rule. So if one needs to do it in faster way then only wa forward I see is the explanation I have mentioned above your post.
To understand Partition Rule, refer to the video attached. _________________
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