Question

In a collection of 6 mathematics books and 4 physics books, the probability that 3 particular mathematics books will be together is:

• A. $\frac{1}{8}$
• B. $\frac{1}{10}$
• C. $\frac{1}{15}$
• D. $\frac{1}{20}$

Answer 1

Answer: (C)

$\frac{1}{15}$

Answer 2

Total ways in a collection of books = (6 + 4)! =10! Let 3 books of maths consider as one book, then (3 maths + 3 maths + 4 physics) = 6 books $\therefore$ Favourable cases = 3! ? 8! $\therefore$ Required probability $= \frac{3 ! \times 8!}{10!} = \frac{3.2.1.8!}{10.9.8!} = \frac{1}{15}$