Question

A bag contains 4 red, 6 white, and 7 blue marbles. How many ways can 3 red, 2 white, and 4 blue marbles be chosen?

Answer 1

Hint: In order to find the total number of ways to choose $ 9 $ marbles containing3 red, 2 white, and 4 blue marbles, assume that all the marbles are identical according to the colour. Since, we have identical marbles in the permutations , use the formula $ \dfrac{{n!}}{{{n_1}!\, \times {n_2}!\, \times .....}} $ where $ n $ is the total number of marbles to permute and $ {n_1},{n_{2,}}{n_{3...}} $ as the number of identical marbles for every colour. Simplify the factorials to obtain the required result.

Complete step-by-step answer:
We are given a bag that contains 4 red, 6 white, and 7 blue marbles in it. In total there are $ 4 + 6 + 7 = 17 $ marbles.
Assuming that all the marbles are identical whether they are red, white or blue.
In this question, we are supposed to find the number of different orders that we could pick the marbles in. That is we are looking for the permutations but as we can see some of the marbles are the same.
According to the question, we are supposed to pick 3 red, 2 white, and 4 blue marbles i.e. total $ 9 $ marbles and if we look for the number of ways by which we can pick 9 marbles , we get the answer as $ 9! $ since there are 9 possibilities for the first marble, times 8 for the second, and so on.
But, here some of these marbles are identical and not different according to colour. Now the formula for permutations in such cases becomes as
 $ \Rightarrow \dfrac{{n!}}{{{n_1}!\, \times {n_2}!\, \times .....}} $
Where $ {n_1} $ is the number of identical objects of one type, $ {n_2} $ is the number of identical objects of another type, and so on.
Here in our case we have a value of $ n = 9 $ i.e. the total number of marbles to permute, in which 3 are of red colour, 2 are of white colour and 4 are of blue colour. Putting all these into the formula mentioned above , we get the number of ways to choose the $ 9 $ marbles as ,
  $ \Rightarrow \dfrac{{9!}}{{3!\, \times 2!\, \times 4!}} $
Simplifying further, we get
 $
   \Rightarrow \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 2 \times 1\, \times 4!}} \\
   \Rightarrow \dfrac{{9 \times 8 \times 7 \times 6 \times 5}}{{3 \times 2 \times 2}} \\
   \Rightarrow 9 \times 4 \times 7 \times 5 \\
   \Rightarrow 1260 \;
  $
Therefore, the total number of ways to choose $ 9 $ marbles is equal to $ 1260 $
So, the correct answer is “ 1260”.

Note: 1. Students be careful while writing the identical number of marbles in the formula.
2. If all the marbles in the bag are different and then we have to find the permutation for $ 9 $ marbles then the answer will be clearly as simple as $ 9! $ .
3. Remember not to skip any step in such questions, as factorial calculation is a little bit confusing and skipping of steps will increase the chance of error in calculations.