How many 6 digit telephone numbers can be formed if each number starts with 91 and sum of the digits is equal to 16?

Options

A. 210
B. 336
C. 168
D. 84

Answer

To solve the problem of how many 6-digit telephone numbers can be formed starting with 91 and having a sum of digits equal to 16, we can follow these steps:

### Step 1: Understand the structure of the telephone number
The telephone number is a 6-digit number that starts with 91. Therefore, the first two digits are fixed as 9 and 1.

### Step 2: Calculate the remaining sum of digits
Since the first two digits (9 and 1) contribute a sum of 10 to the total, we need to find the sum of the remaining four digits such that the total sum equals 16.

So, we have:
\[ 9 + 1 + x_1 + x_2 + x_3 + x_4 = 16 \]
This simplifies to:
\[ x_1 + x_2 + x_3 + x_4 = 16 - 10 = 6 \]

### Step 3: Set up the equation
Now, we need to find the number of non-negative integer solutions to the equation:
\[ x_1 + x_2 + x_3 + x_4 = 6 \]
where \( x_1, x_2, x_3, \) and \( x_4 \) represent the last four digits of the telephone number.

### Step 4: Apply the stars and bars theorem
The stars and bars theorem states that the number of ways to distribute \( n \) indistinguishable objects (stars) into \( r \) distinguishable boxes (bars) is given by the formula:
\[ \binom{n + r - 1}{r - 1} \]

In our case, \( n = 6 \) (the total sum we need) and \( r = 4 \) (the number of digits).

### Step 5: Substitute into the formula
Using the formula, we have:
\[ \binom{6 + 4 - 1}{4 - 1} = \binom{9}{3} \]

### Step 6: Calculate the binomial coefficient
Now we calculate \( \binom{9}{3} \):
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3! \cdot 6!} \]
Calculating this gives:
\[ = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

### Step 7: Conclusion
Thus, the total number of 6-digit telephone numbers that can be formed, starting with 91 and having a sum of digits equal to 16, is **84**.

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