Question

Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 8 green balls and 7 blue balls so that 3 balls of every color are drawn.​

Answer 1

Hint: Here we will first find the number of ways of selecting 3 balls of red color. Then we will find the number of ways of selecting 3 balls of green color and at last we will find the number of ways of selecting 3 balls of blue color. Then we will multiply all these together to get the required number of ways.

Complete step-by-step answer:
Here we need to find the number of ways to draw the 9 balls from a bag such that 3 balls of every color are drawn.
It is given that a bag contains 8 green balls, 6 red balls and 7 blue balls.
Total ball to be drawn \[ = 9\]
So we have to draw 3 balls of each color.
Number of ways of drawing 3 balls of red color \[ = {}^6{C_3}\]
Number of ways of drawing 3 balls of green color \[ = {}^8{C_3}\]
Number of ways of drawing 3 balls of blue color \[ = {}^7{C_3}\]
Total number of ways to draw 9 balls such that 3 balls of every color are drawn will be equal to the product of the number of ways of drawing 3 balls of red color, number of ways of drawing 3 balls of green color and number of ways of drawing 3 balls of blue color.
Required number of ways of drawing the 9 balls \[ = {}^6{C_3} \times {}^8{C_3} \times {}^7{C_3}\].
Using the formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)! \times r!}}\], we get
\[ \Rightarrow \] Required number of ways of drawing the 9 balls \[ = \dfrac{{6!}}{{\left( {6 - 3} \right)! \times 3!}} \times \dfrac{{8!}}{{\left( {8 - 3} \right)! \times 3!}} \times \dfrac{{7!}}{{\left( {7 - 3} \right)! \times 3!}}\]
On further simplification, we get
\[ \Rightarrow \] Required number of ways of drawing the 9 balls \[ = \dfrac{{6!}}{{3! \times 3!}} \times \dfrac{{8!}}{{5! \times 3!}} \times \dfrac{{7!}}{{4! \times 3!}}\]
Now, we will find the values of the factorials.
\[ \Rightarrow \] Required number of ways of drawing the 9 balls \[ = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3 \times 2 \times 1 \times 3!}} \times \dfrac{{8 \times 7 \times 6 \times 5!}}{{5! \times 3 \times 2 \times 1}} \times \dfrac{{7 \times 6 \times 5 \times 4!}}{{4! \times 3 \times 2 \times 1}}\]
On further simplification, we get
\[ \Rightarrow \] Required number of ways of drawing the 9 balls \[ = 5 \times 4 \times 8 \times 7 \times 7 \times 5\]
On multiplying these numbers, we get
\[ \Rightarrow \] Required number of ways of drawing the 9 balls \[ = 39200\]
Hence, the required number of ways of drawing the 9 balls is equal to 39200.

Note: To solve this question, we need to know the basic formulas of permutation and combination. We also need to know the basic difference between the permutation and combination to avoid any mistakes. Permutation is used when we have to find the possible arrangement of elements but the combination is used when we need to find the number of ways to select a number from the collection.