Question

Find the number of groups that can be made from 5 red balls, 3 green balls and 4 black balls, if at least one ball of all colours is always to be included. Given that all balls are identical except colours.

Answer 1

Hint: For the arrangement of objects or numbers, the concept of permutation and combination is used. In the given question, we have to group balls of three types of colours and there should be at least one ball of each colour. So, using the formula to calculate the combination, we can find out the correct answer to the above question.

Complete step-by-step answer:
We have 5 balls of red colour, 3 of green colour and 4 of black colour and we have to select at least one ball of each colour that means there have to be one or more balls of the colours red, green and black. So, the number of possible combinations that can be made under the given conditions is $ N{ = ^5}{C_1}{ \times ^3}{C_1}{ \times ^4}{C_1} $
 $
   \Rightarrow N = \dfrac{{5!}}{{1!\left( {5 - 1} \right)!}} \times \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} \\
   \Rightarrow N = \dfrac{{5 \times 4!}}{{4!}} \times \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{4 \times 3!}}{{3!}} \\
   \Rightarrow N = 5 \times 3 \times 4 \\
   \Rightarrow N = 60 \;
  $
Hence the number of groups that can be made is 60.
So, the correct answer is “60”.

Note: To arrange something in a group in a specific order, we use permutation but when we have to group something and the order of the elements doesn’t matter then we use the concept of combination. In the given question, we are told to form groups that must include at least one ball of each colour specifically, but the order of the balls doesn’t matter that’s why we use the combination.