I'm considering the given prizes as well as the students as distinguishable.
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?
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
p = 4
∴ Total number of 3-digit odd numbers formed= m x n x p = 3 x 5 x 4 = 60.
Permutations & Combinations
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