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Question

Answers

a. 3084080

b. 4084080

c. 5084080

d. 6084080

Answer
Verified

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:-

$5! = 5 \times 4 \times 3 \times 2 \times 1$

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.

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