Aptitude

There are 10 ways of distributing first prize.
There are 10 ways of distributing second prize.
There are 10 ways of distributing third prize.

Three posibilities for distribution of prizes; one person wins all three; two people win all three prizes (two ways); three people win a prize.

Use combination C(n,r)=n!/[r!(n-r)!] n is # in population (10), r is distribution

1st case one winner takes all C(10,1)=10!/[1!*9!]=10

2nd case two winners takes all C(10,2)=10!/[2!*8!]=10*9/2=45
two ways (2,1)&(1,2) 2*C(10,2)=90

3rd case three pople win a prize C(10,3)=10!/[3!*7!]=10*9*8/(3*2)=10*3*4=120

Putting them all together t=C(10,1)+2*C(10,2)+C(10,3)=10+2*45+120=220

Rap

This approach assumes the prizes are indisguishable.

I read the original question as saying three distinct prizes so the answer is 10^3 or 1000.

One could then say that 1st prize is 10, second is 9, and third is 8--so the possibilities is 10*9*8=720.

Rap

But the same person can win multiple prizes so 10x10x10.

There's no mention of distinct. Why would you assume it?