1 votes 1 votes A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw? Verbal Aptitude combinatory + – sushmita asked Mar 30, 2017 • retagged Jun 27, 2017 by Arjun sushmita 1.0k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Akriti sood commented Mar 30, 2017 reply Follow Share 3C1 *6C2 + 3C2 * 6C1 + 3C3 * 6C0 2 votes 2 votes sushmita commented Mar 30, 2017 reply Follow Share why are we considering balls as distinct?? 0 votes 0 votes Akriti sood commented Mar 31, 2017 reply Follow Share i did nt consider thm as different.i am using combination,not permutation 0 votes 0 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes At least one black ball is to be included, So there can be three cases, (1 black ball $ \& $ 2 other ball) + (2 black ball $ \& $ 1 other ball) + (3 black ball $ \& $ 0 other ball) $( 3C_1 * 6C_2 ) + ( 3C_2 * 6C_1 ) + ( 3C_3 * 6C_0 )$ simrankapoor8 answered Mar 30, 2017 • selected Apr 2, 2017 by rude simrankapoor8 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw? 2 white balls 3 black balls 4 red balls. white black red (atleat one black ball) 0 3 0 0 2 1 1 2 0 1 1 1 3C3 + 2C2 * 4C1 + 2C1*3C2 +2C1 * 3C1 * 4C1=1 + 4 + 6 +24=35 akankshadewangan24 answered May 4, 2017 akankshadewangan24 comment Share Follow See all 0 reply Please log in or register to add a comment.