\(P+Q+R+S = 7\)

\(P_{min} = 1\)
\(Q_{min} = 1\)

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



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


_________________

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
_________________

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:)

Sir can you explain this highlighted part. Why we put 3 partition in 5. When there are 4 boxes.

Posted from my mobile device

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
_________________

Can you please explain this a bit. Why are you doing 5+3?

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

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
_________________

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

dave13

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.
_________________

Hi everyone!

I fail to understand why all solutions only include 1 ball in each of P and R. To my mind, we should also count all cases when there are more balls in them ,i.e. total number of cases would be 3C10 - 1C8 = 112

Bunuel chetan2u - would really appreciate your comments, thank you!