Question

There are 9 different colors of paint to choose from. Out of the 9 colors, how many ways can 4 different colors be chosen?

Answer 1

Hint:Draw a table with four columns, and then imagine in how many ways you can fill the first box if you have “n” number of choices? Yes, if you have “n” number of choices then you can fill the first box in “n” number of ways, so now you are left with “n-1” number choices, so similarly you will have “n-1” number of ways to fill the next box. Use this logic to solve the question.

Complete step by step solution:
In order to find in how many ways four different colors can be chosen from nine different colors, we will first create a table consist of four columns

\[{1^{st}}\]

\[{2^{nd}}\]

\[{3^{rd}}\]

\[{4^{th}}\]

Now, for the first box we have $9$ different colors to choose, so we can fill the first box in $9$ different ways.

\[{1^{st}} =9\]

\[{2^{nd}}\]

\[{3^{rd}}\]

\[{4^{th}}\]

Now, we have $8$ different colors left because one out of nine is being filled in the first box, so we can fill second box in $8$ different ways and similarly, we can fill third and forth box in $7\;{\text{and}}\;6$ different ways respectively.

\[{1^{st}} =9\]

\[{2^{nd}} =8\]

\[{3^{rd}} =7\]

\[{4^{th}} =6\]

And if we talk about the total number of ways to fill or choose four different colors from nine different colors, this will be equal to $9 \times 8 \times 7 \times 6 = 3024$ ways.

Note: Creating a table for this type of questions is proved to be very helpful when imagining the situation. In actual, one needs not to imagine situation or solve this question practically, if he knows that it can be solved in one more way, since this question is a problem of permutation and we can directly solve it as:
\[{}^9{{\text{P}}_4} = \dfrac{{9!}}{{(9 - 4)!}} = \dfrac{{9!}}{{5!}} = \dfrac{{1 \times 2 \times 3 \times 4
\times 5 \times 6 \times 7 \times 8 \times 9}}{{1 \times 2 \times 3 \times 4 \times 5}} = 6 \times 7\times 8 \times 9 = 3024\]