Manipal Entrance Test 2014 Math Paper

There may be two cases.
Case I. Two children get none and one get three and others get one each.
Then, total number, of ways =

10!

2!×3!×7!

×10!
Case II. Two get none and two get 2 each and the others get one each.
Then, total number of ways =

10!

(2!)4×6!

×10!
Hence, total number of ways =

(10!)2

2!×3!×7!

+

(10!)2

(2!)4×6!

=

(10!)2×25

(2!)2×6!×84