The number of ways of selecting 8 books from a library which has 9 books each of Mathematics, Physics, Chemistry and English is:

Options

A. 165
B. 166
C. 167
D. 168

Answer

To solve the problem of selecting 8 books from a library that has 9 identical books each of Mathematics, Physics, Chemistry, and English, we can use the "stars and bars" theorem from combinatorics. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We need to select a total of 8 books from 4 different subjects (Mathematics, Physics, Chemistry, and English), where each subject has an unlimited supply of identical books (up to 9 in this case).

### Step 2: Define Variables
Let:
- \( x_1 \) = number of Mathematics books selected
- \( x_2 \) = number of Physics books selected
- \( x_3 \) = number of Chemistry books selected
- \( x_4 \) = number of English books selected

We need to find the number of non-negative integer solutions to the equation:
\[ x_1 + x_2 + x_3 + x_4 = 8 \]

### Step 3: Apply the Stars and Bars Theorem
According to the stars and bars theorem, the number of ways to distribute \( n \) identical items (stars) into \( r \) distinct groups (bars) is given by:
\[ \binom{n + r - 1}{r - 1} \]

In our case:
- \( n = 8 \) (the total number of books to select)
- \( r = 4 \) (the number of subjects)

### Step 4: Substitute Values into the Formula
Now we can substitute the values into the formula:
\[ \binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} \]

### Step 5: Calculate the Binomial Coefficient
Now we need to calculate \( \binom{11}{3} \):
\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \]

### Conclusion
Thus, the number of ways to select 8 books from the library is \( 165 \).