Each of 10 passengers boards any of the three buses randomly, which had no passenger initially. The probability that each bus has got atleast one passenger is

n(S) = No.of ways 10 passengers can board 3 busses = 3¹⁰
No.of favourable cases = (No. of surjections from A to B where n(A) =10, n(B) =3)
$=\left[3^{10}{-}^3{\mathrm{C}}_1\left(2^{10}\right){+}^3{\mathrm{C}}_2(1{)}^{10}\right]$
$\therefore$Required probability $=\left[\frac{3^{10}-3(2{)}^{10}+3}{3^{10}}\right]$