Question

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

Answer 1

Hint: If we say at least one black ball it means we only need to consider cases in which at least one black ball in that group.
In the general number of ways to select r things from n number of things is ${}^{n}{{C}_{r}}$.
In general we can expand ${}^{n}{{C}_{r}}$as
$^{n}{{C}_{r}}=\dfrac{n!}{r!\times (n-r)!}$

Complete step by step solution:
Total number of balls in box = 2 white balls + 3 black balls + 4 red balls = 9
First we can find all possible ways to select 3 balls from 9 balls i.e
${{\Rightarrow }^{9}}{{C}_{3}}=\dfrac{9!}{3!\times (9-3)!}$
${{\Rightarrow }^{9}}{{C}_{3}}=\dfrac{9!}{3!\times 6!}$
${{\Rightarrow }^{9}}{{C}_{3}}=\dfrac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}$
${{\Rightarrow }^{9}}{{C}_{3}}=84$
Now we can find total possible ways of selecting 3 balls from total 6 balls( 2 white balls + 4 red balls) we can also say total non black balls i.e
${{\Rightarrow }^{6}}{{C}_{3}}=\dfrac{6!}{3!\times (6-3)!}$
${{\Rightarrow }^{6}}{{C}_{3}}=\dfrac{6!}{3!\times 3!}$
${{\Rightarrow }^{6}}{{C}_{3}}=\dfrac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}$
${{\Rightarrow }^{6}}{{C}_{3}}=20$
Hence the total number of ways to draw three balls from a box in which at least one black ball is = 84-20=64.

Note: As we only need to choose the cases in which we should have one black ball that’s why we need to remember to subtract total number of ways to select 3 balls and total number of ways to select three balls in which there is no black ball.
We can expand factorial of n as
$n!=n\times (n-1)\times (n-2)\times (n-3)\times ..............\times 4\times 3\times 2\times 1$