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

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

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


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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:)

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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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

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