If a,b,c and d are odd natural numbers such that a+b+c+d=20, the number of values of the ordered quadruplet (a,b,c,d) is

Options

A. 165
B. 455
C. 310
D. None of these

Explanation

To solve the problem, we need to find the number of ordered quadruplets (a, b, c, d) of odd natural numbers such that their sum equals 20.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that a, b, c, and d are odd natural numbers. The sum of these numbers is given as:
\[ a + b + c + d = 20 \]

2. **Expressing Odd Numbers**:
Any odd natural number can be expressed in the form:
\[ a = 2n_1 + 1, \quad b = 2n_2 + 1, \quad c = 2n_3 + 1, \quad d = 2n_4 + 1 \]
where \( n_1, n_2, n_3, n_4 \) are non-negative integers.

3. **Substituting into the Equation**:
Substitute the expressions for a, b, c, and d into the equation:
\[ (2n_1 + 1) + (2n_2 + 1) + (2n_3 + 1) + (2n_4 + 1) = 20 \]
Simplifying this gives:
\[ 2(n_1 + n_2 + n_3 + n_4) + 4 = 20 \]

4. **Rearranging the Equation**:
Now, isolate the term involving \( n \):
\[ 2(n_1 + n_2 + n_3 + n_4) = 20 - 4 \]
\[ 2(n_1 + n_2 + n_3 + n_4) = 16 \]
Dividing both sides by 2:
\[ n_1 + n_2 + n_3 + n_4 = 8 \]

5. **Finding the Number of Solutions**:
We need to find the number of non-negative integer solutions to the equation:
\[ n_1 + n_2 + n_3 + n_4 = 8 \]
The formula for the number of solutions in non-negative integers for the equation \( x_1 + x_2 + ... + x_r = n \) is given by:
\[ \binom{n + r - 1}{r - 1} \]
Here, \( n = 8 \) and \( r = 4 \):
\[ \text{Number of solutions} = \binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} \]

6. **Calculating the Binomial Coefficient**:
Now, we calculate \( \binom{11}{3} \):
\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \]

### Final Answer:
Thus, the number of values of the ordered quadruplet (a, b, c, d) is:
\[ \boxed{165} \]