In how many ways 8 students can be given 3 prizes such that no student receives more than 1 prize

I'm considering the given prizes as well as the students as distinguishable.

Índice Show

5. none of these

First we make from the given prizes $5$ nonempty packages. Their sizes can be $(3,1,1,1,1)$ or $(2,2,1,1,1)$. In the first case we can form the $3$-pack in ${7\choose3}=35$ ways, in the second case we can form two $2$-packs in ${1\over2}{7\choose2}{5\choose2}=105$ ways.

We now have $5$ different packages that can be dealt to the students in $5!$ ways. The total number of choices then is $$N=(35+105)\cdot5!=16\,800\ .$$ (In certain cases $N$ could be smaller: The given prizes could be six balls of colors blue, red, white, yellow, black, and orange, and a package containing three balls of colors blue, red, white.)

How many 3-digit odd numbers can be formed from the digits 1,2,3,4,5,6 if:
(a) the digits can be repeated (b) the digits cannot be repeated?

(a) Number of digits available = 6

Number of places [(x), (y) and (z)] for them = 3Repetition is allowed and the 3-digit numbers formed are oddNumber of ways in which box (x) can be filled = 3 (by 1, 3 or 5 as the numbers formed are to be odd)

m = 3

Number of ways of filling box (y) = 6 (∴ Repetition is allowed)

n = 6

Number of ways of filling box (z) = 6 (∵ Repetition is allowed)

p = 6

∴ Total number of 3-digit odd numbers formed = m x n x p = 3 x 6 x 6 = 108(b) Number of ways of filling box (x) = 3 (only odd numbers are to be in this box )

m = 3

Number of ways of filling box (y) = 5 (∵ Repetition is not allowed)

n = 5

Number of ways of filling box (z) = 4 (∵ Repetition is not allowed)

p = 4

∴ Total number of 3-digit odd numbers formed

= m x n x p = 3 x 5 x 4 = 60.