STORY OF 5 MARBLES

We conduct the experiment below. The rule of the game is that

(1) All marbles are to be allocated to boxes.

(2) Blank boxes are allowed.

(3) There is no restriction on number of marbles that can be put in any box.

(4) Though here marble no. and no. of boxes are same, they can be different and formula

will apply. Keep n > r

 No. of Boxes(r): 1 No. of Marbles (n): 1 No. of Arrangement: C(n+r-1,n) 1 No. of Non-Empty Arrangement: C(n-1,n-r) 1 Details of Arrangement- Box 1 1 arrangement(1) Formula- C(1,1) •

 No. of Boxes (r): 2 No. of Marbles (n): 2 No. of Arrangement: C(n+r-1,n) 3 No. of Non-Empty Arrangement: C(n-1,n-r) = C(n-1,r-1) 1 Details of Arrangement: Box 1 Box 2 1 arrangement (1+1) Formula-C(2,2) • • 2 arrangements(2+0) Formula- C(2,1) •• ••

 No. of Boxes (r): 3 No. of Marbles (n): 3 No. of Arrangement: C(n+r-1,n) 10 No. of Non-Empty Arrangement: C(n-1,n-r) 01 Details of Arrangement: Box 1 Box 2 Box 3 3 arrangements(3+0)ormula-C(3,1) ••• ••• ••• 1 arrangement(1+1+1) Formula- C(3,3) • • • 6 arrangements(2+1) Formula-C(3,1)*C(2,1) •• • •• • • •• •• • • •• • ••

 No. of Boxes (r): 4 No. of Marbles (n): 4 No. of Arrangement: C(n+r-1,n) 35 No. of Non-Empty Arrangement: C(n-1,n-r) 01 Details of Arrangement: Box 1 Box 2 Box 3 Box 4 1 arrangement(1+1+1+1) Formula-C(4,4) • • • • 4 arrangements(4+0) Formula- C(4,1) •••• •••• •••• •••• 12 arrangements(3+1) Formula - C(4,1)*C(3,1) ••• • ••• • ••• • • ••• ••• • ••• • • ••• • ••• ••• • • ••• • ••• • ••• 6 arrangements(2+2) Formula- C(4,2) •• •• •• •• •• •• •• •• •• •• •• •• 12 arrangements(2+1+1) Formula- C(4,1)*C(3,2) •• • • •• • • •• • • • •• • • •• • •• • • • • •• • •• • • •• • • • •• • • •• • • ••

 No. of Boxes (r): 5 No. of Marbles (n): 5 No. of Arrangement: C(n+r-1,n) 126 No. of Non-Empty Arrangement: C(n-1,n-r) 01 Details of Arrangement: Box 1 Box 2 Box 3 Box 4 Box 5 1 arrangement(1+1+1+1+1) Formula- C(5,5) • • • • • 5 arrangement (5+0) Formula- C(5,1) ••••• ••••• ••••• ••••• ••••• 20 arrangements (4+1) Formula- C(5,1)*C(4,1) •••• • •••• • •••• • •••• • • •••• •••• • •••• • •••• • • •••• • •••• •••• • •••• • • •••• • •••• • •••• •••• • • •••• • •••• • •••• • •••• 20 arrangements(3+2) Formula- C(5,1)*C(4,1) ••• •• ••• •• ••• •• ••• •• •• ••• ••• •• ••• •• ••• •• •• ••• •• ••• ••• •• ••• •• •• ••• •• ••• •• ••• ••• •• •• ••• •• ••• •• ••• •• ••• 30 arrangements(3+1+1) Formula- C(5,1)*C(4,2) ••• • • ••• • • ••• • • ••• • • ••• • • ••• • • • ••• • • ••• • • ••• • ••• • • ••• • • ••• • • • • ••• • ••• • • ••• • • ••• • ••• • • • ••• • • • ••• • • ••• • ••• • • ••• • • ••• • • • ••• • • ••• • • ••• • • ••• • • ••• • • ••• • • ••• 30 arrangements (2+2+1) Formula- C(5,2)*C(3,1) •• •• • •• •• • •• •• • • •• •• •• •• • •• •• • • •• •• • •• •• •• •• • • •• •• • •• •• • •• •• •• • •• •• • •• •• • •• •• • •• •• • •• •• •• • • •• •• •• • •• •• •• • • •• •• •• • •• •• • •• • •• •• • •• •• •• • •• •• • •• •• •• • •• •• • 20 arrangements (2+1+1+1) Formula-C(5,1)*C(4,3) •• • • • •• • • • •• • • • •• • • • • •• • • • •• • • • •• • • •• • • • • • •• • • • •• • • •• • • • •• • • • • • •• • • •• • • • •• • • • •• • • • • •• • • • •• • • • •• • • • ••

5 boxes and 5 marbles

 No. of empty boxes Configuration No. of Arrangements Probability zero 1+1+1+1+1 01 1/126 1 2+1+1+1 20 20/126 2 ( Formula- 2*C(5,1)*C(4,2) ) 3+1+1 30 60/126 2+2+1 30 3 ( Formula- 2*C(5,1)*C(4,1) ) 4+1 20 40/126 3+2 20 4 5+0 05 05/126

* What is the probability that exactly 2 boxes are empty ?   60/126

* What is the probability that not more than 2 boxes are empty ?   81/126

* What is the probability that at least 2 boxes are empty ? 60/126 +40/126 +05/126= 105/126

* What is the probability that no box is empty ? 1/126

n Locks and n Keys

 Suppose there are n locks and n keys. What is the minimum no. of trials in which Lock-key will be matched and what is the maximum no. of trials ? The problem can be extended to above problem of marbles if  marbles & boxes are numbered and each marble is assigned a particular box. Ans: In case of minimum no. of trials, 1st key will be matched with 1st lock in first trial.; 2nd key with 2nd lock in 2nd trial; nth key with nth lock in nth trial. Hence no. of  minimum trials- nIn case of maximum no. of trials, 1st key will match with 1st lock on n no. of trials, 2nd with 2nd lock in n-1 no. of trials, ........ last key with last lock in single trial because only 1 key and 1 lock will be left. Hence maximum no. of trials n+(n-1)+(n-2)+.......+3+2+1 = n(n+1)/2 =C(n+1,2) * Average no. of trials is n(n+3) / 4. * No. of Maximum trials is (n+1)/2 times the minimum no. of trials and hence proportional to n.