**Question:**

In how many ways can 4 prizes be distributed among 5 students, when

(i) no student gets more than one prize?

(ii) a student may get any number of prizes?

(iii) no student gets all the prizes?

**Solution:**

(i) Since no student gets more than one prize; the first prize can be given to any one of the five students.

The second prize can be given to anyone of the remaining 4 students.

Similarly, the third prize can be given to any one of the remaining 3 students.

The last prize can be given to any one of the remaining 2 students.

$\therefore$ Required number of ways $=5 \times 4 \times 3 \times 2=5 !$

(ii) Since a student may get any number of prizes, the first prize can be given to any of the five students. Similarly, the rest of the three prizes can be given to the each of the remaining 4 students.

$\therefore$ Required number of ways $=5 \times 5 \times 5 \times 5=625$

(iii) None of the students gets all the prizes.

$\therefore$ Required number of ways $=\{$ Total ways of distributing the prizes in a condition wherein a student may get any number of prizes $-$ Total ways in a condition in which a student receives all the prizes $\}=$

ways in a condition in which a student receives all the prizes $\}=$ $625-5=620$