A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from a box if atleast one black ball is to be included in the draw?

\l|ccc White Black Red \ hline 2 3 4 \ bfCase 1: 0 1 2 \ 2 1 0 \ 1 1 1 \ bfCase 2: 0 2 1 \ 1 2 0 \ bfCase 3: 0 3 0 \ = ^3C_1( ^4C_2 + ^2C_2 + ^2C_1 x ^4C_1 ) + ^3C_2( ^4C_1 + ^2C_1 ) + ^3C_3 = 64 Number of ways of selecting 3 balls with at least one black ball = Total number of selections - Number of selections with no black ball = ^9C_3 - ^6C_3 = 84 - 20 = 64.

A box contains two white balls, three black balls and f... | Permutations and Combinations (Mathematics)

\begin{array}{l|ccc} & \text{White} & \text{Black} & \text{Red} \\ \hline & 2 & 3 & 4 \\ \textbf{Case 1:} & 0 & 1 & 2 \\ & 2 & 1 & 0 \\ & 1 & 1 & 1 \\ \textbf{Case 2:} & 0 & 2 & 1 \\ & 1 & 2 & 0 \\ \textbf{Case 3:} & 0 & 3 & 0 \end{array}

\\= \ ^{3}C_{1}\left( \ ^{4}C_{2} + \ ^{2}C_{2} + \ ^{2}C_{1} \times \ ^{4}C_{1} \right) + \ ^{3}C_{2}\left( \ ^{4}C_{1} + \ ^{2}C_{1} \right) + \ ^{3}C_{3} = 64 \\

Number of ways of selecting 3 balls with at least one black ball = Total number of selections - Number of selections with no black ball

\\= \ ^{9}C_{3} - \ ^{6}C_{3} = 84 - 20 = 64

Explanation

For (2 leq r leq n,binom{n}{r} + 2binom{n}{r - 1} + binom{n}{r - 2} =)

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