Question

An urn contains $5$ red, $4$ black and $3$ white marbles. Then the number of ways in which $4$ marbles can be drawn from it so that at most $3$ of them are red, is:

• A. $455$
• B. $460$
• C. $490$
• D. $495$

Answer 1

Answer: (C)

$490$

Total marbles are $5$ red, $4$ black and $3$ white.

$4$ marbles are drawn, atmost $3$ of them are red.

Case are:

$3R$ and $1$ other ${\to }^5{C}_3{\times }^7{C}_1=70$

$2R$and $2$ other ${\to }^5{C}_2{\times }^7{C}_2=210$

$1R$ and $3$ other ${\to }^5{C}_1{\times }^7{C}_3=175$

Zero $R$ and $4$ other ${\to }^5{C}_0{\times }^7{C}_4=35$

Toal $=490$

Correct option is C.