Question

Find the number of non-negative solutions of the system of equations: $a+b=10,a+b+c+d=21,a+b+c+d+e+f=33,a+b+c+d+e+f+g+h=46,....,a+b+...+y+z=208.$

• A. $P_{10}^{22}$
• B. $P_{11}^{22}$
• C. $P_{13}^{23}$
• D. None of these.

Answer 1

The correct option is C

$P_{13}^{23}$

Step:1 Derive the general form of non-negative integral solutions in an equation.

Let, $a_1+a_2+a_3+a_4.........a_n=x_1$ is an equation.

then number of non-negative integral solutions $=C_{n-1}^{x_1+n-1}$, where $n=$number of terms in that equation.

Step:2 Calculating the number of non-negative integral solutions

1.Calculating value for first equation

For the equation $a+b=10$, number of non-negative integral solutions

$$\begin{gathered} =C_{2-1}^{10+2-1} \\ =C_1^{11} \\ =11 \end{gathered}$$

2.Calculating value for second equation

Given equation is $a+b+c+d=21$.

Therefore, $c+d=11$, since $a+b=10$

For the equation $c+d=11$, number of non-negative integral solutions

$$\begin{gathered} =C_{2-1}^{11+2-1} \\ =C_1^{12} \\ =12 \end{gathered}$$

3. Calculating value for last equation

Similarly, we can write $y+z=22$ from the given equation $a+b+...+y+z=208.$

Now, for the equation $y+z=22$, number of non-negative integral solutions

$$\begin{gathered} =C_{2-1}^{22+2-1} \\ =C_1^{23} \\ =23 \end{gathered}$$

Step:3 Calculating the number of non-negative integral solutions for the entire system

Therefore, for the system containing equations: $a+b=10,a+b+c+d=21,a+b+c+d+e+f=33,a+b+c+d+e+f+g+h=46,....,a+b+...+y+z=208.$ is:

$$\begin{gathered} 11\times 12\times ......\times 23 \\ =P_{11}^{23} \end{gathered}$$

Therefore, Option(C) is the correct answer.