If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans In how many ways can

1-(7c7*5c1+5c5*7c3)/12c8=455
10 years agoHelpfull: Yes(14) No(4)
7c3*5c5+7c4*5c4+7c5*5c3+7c6*5c2+7c7*5c1
= 7*5 + 35*5 + 21*10 + 70 + 5
= 35 + 175 + 210 + 75
=495
10 years agoHelpfull: Yes(10) No(11)
draw 8 balls contains atleast 1 blue and 1 red ball are
(7B AND 1R)OR(6B AND 2R)OR(5B AND 3R)OR(4B AND 4R)OR(3B AND 5R)
=(7c7*5c1)*(7c6*5c2)*(7c5*5c3)*(7c4*5c4)*(7c3*5c5)
=495
10 years agoHelpfull: Yes(7) No(9)
5+70+210+175+35=495
10 years agoHelpfull: Yes(4) No(6)
we have 3 chances;(6b,2r);(5b,3r);(4b,4r);
total no of ways=(7c6*5c2)+(7c5*5c3)+(7c4*5c4)
=(7*10)+(21*10)+(35*5)
=455
10 years agoHelpfull: Yes(4) No(1)
Possible ways to draw 8 balls from the refrigerator which contains atleast 1 blue and 1 red can after the drawing are (6,2) (5,3) (4,4).
For (6, 2) = ⇒7c6*5c2⇒7*10=70
For (5, 3) = ⇒7c5*5c3⇒21*10=210
For (4, 4) = ⇒7c4*5c4⇒35*5=175
So Total ways = 70+210+175=455
9 years agoHelpfull: Yes(2) No(1)
280+560+350=1190
10 years agoHelpfull: Yes(1) No(11)
it vl be 455
10 years agoHelpfull: Yes(1) No(3)

If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that atleast 1 blue can and 1 red can remains in the refrigerator.