Question

In how many ways can 15 billiard balls be arranged in a row if 3 are red, 4 are white and 8 are black?

Answer 1

Solution:

→Number of Billiard balls = 15 balls [ 3 Red  + 4 White + 8 Black]

→There are 3 red balls of same type .

→4 White balls are of same type .

→ And 8 Black balls are of same type .

→3 Red (R) , 4 White (W), and 8 Black (B) can arranged in line i.e in a row is given by=

→ Arrangements of 15 things in which 3 is of a kind, 4 is of different kind and 8 is of another kind.

                                           = $\frac{15!}{3!\times4!\times8!}$

  As, 15! = 15 × 14 ×13×12×11×10×9×8!

and, 4! = 4 × 3×2=24  ∧  3! = 3 × 2 × 1 = 6              

→   $\frac{15!}{8!}$ = $\frac{15\times 14 \times13\times12\times11\times10\times9\times8! }{8!}$ =15 × 14×13×12×11×10= 25 ×11×13×7×144

And , 4! × 3! = 24 × 6 = 144

=$=\frac{25\times11\times13\times7\times144}{4!\times3!}\\ =\frac{25\times11\times13\times7\times144}{24\times6} \\=  25 \times 13\times 11\times 7$

= 25025