Question

In how many ways can 17 billiard balls be arranged, if 7 of them are black, 6 red and 4 white?
a. 3084080
b. 4084080
c. 5084080
d. 6084080

Answer 1

Hint: To solve this question we will first assume that the balls of same colour are identical which means their physical appearance, weight etc. are similar. Now to arrange these balls we will require a following formula i.e.
Total arrangement$ = \dfrac{{X!}}{{({x_1}!)({x_2}!)({x_3}!).........({x_n}!)}}$
Here $X!$ is the total number of entities, $({x_1})$ is an entity of the same colour but different from$({x_2}),({x_3})$.So all three are different from each other.
Also we should know how to solve a factorial. For example if we want to calculate factorial of 5 then it will be solved as follows:-
$5! = 5 \times 4 \times 3 \times 2 \times 1$

Complete step-by-step answer:
Total ways of arranging 17 billiard balls having (7B, 6R, 4W)$ = \dfrac{{X!}}{{({x_1}!)({x_2}!)({x_3}!).........({x_n}!)}}$
With the help of this formula we can find the total number of ways where 17 billiards balls can be arranged.
${x_1}! = $Number of ways black balls can be arranged
${x_2}! = $Number of ways red balls can be arranged
${x_3}! = $Number of ways white balls can be arranged
$X! = $Total ways possible for 17 billiard balls.
$ = \dfrac{{17!}}{{7! \times 6! \times 4!}}$
$ = \dfrac{{17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7!}}{{7! \times 6! \times 4!}}$ (first expanding factorial of 17)
$ = \dfrac{{17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 4 \times 3 \times 2 \times 1}}$ (Expanding factorial 6 and 4, canceling factorial 7)
$ = \dfrac{{17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}}{{(6 \times 2)(4 \times 4)(5 \times 2)(3 \times 3)}}$ (Making denominator terms like numerator so that they can be cancelled)
$ = \dfrac{{17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}}{{(12)(16)(10)(9)}}$ (Cancelling equal terms from numerator and denominator)
$ = 17 \times 15 \times 14 \times 13 \times 11 \times 8$
$ = 4084080$
Thus there are $4084080$ ways in which we can arrange these billiard balls. So, option (B)
Is correct.

Note: We cannot arrange the total number of 17 billiard balls only as $17!$ .It is only valid when all the balls are distinct, not identical but here in our case all the balls of same colour are identical so this cannot be applied. Most of the students make this mistake quite often so do cautiously.