Question

Q13.       Let S be a finite set containing 12 elements, which we wish to partition into Cells C1, C2, C3, and C4, such that n(C1) = 2, n(c2) = 2, n(C3) = 3, n(C4) = 5.

How many such partitions are possible?

Answer 1

Number of ways in which n items can be selected from r, nCr = n!/(r! x (n-r)!)

Number of ways in which first 2 elements can be selected for C1 = 12C2

= 12!/(2! x 10!)

= 66

Number of ways in which 2 elements can be selected for C2 from remaining 10 = 10C2

= 10!/(2! x 8!)

= 45

Number of ways in which 3 elements can be selected for C3 from remaining 8 = 8C3

= 8!/(5! x 3!)

= 56

Number of ways in which 5 elements can be selected from remaining 5 = 5C5

= 1

So, number of partitions possible = 66 x 45 x 56 x 1

= 166,320