Let U be all possible distributions of or distinct object into 12 boxes into 12 boxes. Let Ai denote the set of possible distributions with the set of possible distributions with the ith box being empty.
Then the required number of distribution with at least one empty box is |A1 U A2 U ………..U A12| We have N = 12r. Also, |Ai| = (12 - 1)r , the number of distribution in witch the objects are put into of the remaining four boxes.
Similarly, |Ai ∩ Aj| = (12 - 2)r , and so eleventh. Thus by the corollary above , we have
|A1 U A2 U ……… U A12| = S1 – S2 + S3 – S4 + S5 – S6 + S7 – S8 + S9 – S10 + S11 – S12
= C (12,1)11r – C(12,2)10r + C(12,3)9r - C (12,4)8r + C(12,5)7r - C(12,6)6r + C (12,7)5r – C(12,8)4r + C(12,9)3r - C (12,10)2r + C(12,11)1r – 0
ANSWER = No empty box