Question

A bag contains $30$ tokens numbered serially from $0$ to $29.$ The number of ways of choosing $3$ tokens from the bag, such that the sum on them is $30,$ is

• A. $56$
• B. $44$
• C. $75$
• D. $81$

Answer 1

Answer: (C)

Let, $a,b,c$ be the selected numbers; $a+b+c=30;a < b < c.$
$a=0\Rightarrow \left(b , c\right)=\left(1,29\right)\left(2,28\right).....\left(14,16\right)$ $ \rightarrow Total14ways.$
$a=1\Rightarrow \left(b , c\right)=\left(2,27\right)\left(3,26\right).....\left(14,15\right)$ $ \rightarrow Total13ways.$
$a=2\Rightarrow \left(b , c\right)=\left(3,25\right)\left(4,24\right).....\left(13,15\right)$ $ \rightarrow Total11ways.$
$a=3\Rightarrow \left(b , c\right)=\left(4,23\right)\left(5,22\right).....\left(13,14\right)$ $ \rightarrow Total10ways.$
$a=4\Rightarrow \left(b , c\right)=\left(5,21\right)\left(6,20\right).....\left(12,14\right)$ $ \rightarrow Total8ways.$
$a=5\Rightarrow \left(b , c\right)=\left(6,19\right)\left(7,18\right).....\left(12,13\right)$ $ \rightarrow Total7ways.$
$a=6\Rightarrow \left(b , c\right)=\left(7,17\right)\left(8,16\right).....\left(11,13\right)$ $ \rightarrow Total5ways.$
$a=7\Rightarrow \left(b , c\right)=\left(8,15\right)\left(9,14\right).....\left(11,12\right)$ $ \rightarrow Total4ways.$
$a=8\Rightarrow \left(b , c\right)=\left(9,13\right)\left(10,12\right) \rightarrow Total2ways.$
$a=9\Rightarrow \left(b , c\right)=\left(10,11\right) \rightarrow Total1way.$
So, the total number of ways $=1+2+4+5+7+8+10+11+13+14=75.$