There are 12 holes made in the ground. At least 3 are to be filled with a red ball and the other holes can be filled with any color ball. In how many different ways can all the holes be filled from a box of 5 red balls and 10 mixed color balls?

3 Red => 9 coloured => $$^{10}C_9$$ * 12!/3! ways = 10*12*11*10/6 = 1100*2 = 2200
4 Red => 8 coloured => $$^{10}C_8$$ * 12!/4! ways = 45*12*11*10*9/24 = 22275
5 Red => 7 coloured => $$^{10}C_7$$ * 12!/5! ways = 479001600

Total number of ways = 479026075

Hi Aaditya, thanks for the response! :)
But this wasn't in the options. the options were
345
425
445
465
485
with 445 being the answer.

[12! / (3! * 9!) ] [ ( (5C3 * 10C9) + (5C4 * 10C8) + (5C5 * 10C7) ) / 12C3 ]

this one is very easy abhipraay .
look :
5c3x10c9=100
5c4x10c8=225
5c5x10c7=120
Sum of all three= 445 .
Hope you get it :) .. we had a condition of atleast 3 there can be more than 3 red balls however since there are only 5 balls we have only these possibilities . :)

Hey Gaurav, you're answer is correct however, I have 2 Questions.
1.) By the same logic, why can't I use 5c3 X 12c9.
2.) Why are you not considering combination of arrangement of these balls in 12 holes