Question

There are stalls for 10 animals in a ship.The number of ways in which ship load can be made, if there are cows , calves and horses to be transported, animals of each kind being not less than 10, is:

Answer 1

Using multinomial theorem, we have

$x_1+x_2+x_3=10$

$0\le x_1\le 10$ = no. of coues transported

$0\le x_2\le 10$ = No. of calues transported

$0\le x_3\le 10$ = No. of horses transported

$\Rightarrow$ No. of ways is ${}^{n+r-1}{C}_r$

n = 3 r = 10

${\Rightarrow }^{n+r-1}{C}_r{=}^{10+5-1}{C}_{10}$

${=}^{12}{C}_{10}=66$

$\therefore$ there are 66 ways