Q.17. Let A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2c = 22}. The n(A) is equal

Q: What is the correct answer for JEE Main Mathematics Q17?

The correct answer is 3. Sol. We need to find the number of non-negative integer solutions to $a + b + 2c = 22$. We can iterate on the possible values of $c$. Since $a, b \ge 0$, we must have $2c \le 22$, so $c \le 11$. Case 1: $c=0 \Rightarrow a+b = 22$. Number of non-negative solutions is ${22+2-1 \choose 2-1} = {23 \choose 1} = 23$. Case 2: $c=1 \Rightarrow a+b = 20$. Number of non-negative solutions is ${20+2-1 \choos

Q: Q.17. Let A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2c = 22}. The n(A) is equal to :

The correct answer is 3. Sol. We need to find the number of non-negative integer solutions to $a + b + 2c = 22$. We can iterate on the possible values of $c$. Since $a, b \ge 0$, we must have $2c \le 22$, so $c \le 11$. Case 1: $c=0 \Rightarrow a+b = 22$. Number of non-negative solutions is ${22+2-1 \choose 2-1} = {23 \choose 1} = 23$. Case 2: $c=1 \Rightarrow a+b = 20$. Number of non-negative solutions is ${20+2-1 \choos

Q: What are the options for this JEE Main Mathematics question?

A. 121 · B. 124 · C. 144 · D. 169

Question

Q.17. Let A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2c = 22}. The n(A) is equal to :

FirstInTest · Accepted Answer

Sol. We need to find the number of non-negative integer solutions to $a + b + 2c = 22$.
We can iterate on the possible values of $c$.
Since $a, b \ge 0$, we must have $2c \le 22$, so $c \le 11$.
Case 1: $c=0 \Rightarrow a+b = 22$. Number of non-negative solutions is ${22+2-1 \choose 2-1} = {23 \choose 1} = 23$.
Case 2: $c=1 \Rightarrow a+b = 20$. Number of non-negative solutions is ${20+2-1 \choose 2-1} = {21 \choose 1} = 21$.
Case 3: $c=2 \Rightarrow a+b = 18$. Number of non-negative solutions is ${18+2-1 \choose 2-1} = {19 \choose 1} = 19$.
...
Case $k$: $c=k \Rightarrow a+b = 22-2k$. Number of non-negative solutions is ${22-2k+2-1 \choose 2-1} = {23-2k \choose 1} = 23-2k$.
...
Case 11: $c=11 \Rightarrow a+b = 0$. Number of non-negative solutions is ${0+2-1 \choose 2-1} = {1 \choose 1} = 1$.
The total number of solutions is the sum of these counts:
$N = 23 + 21 + 19 + ... + 1$.
This is an arithmetic progression with first term $a_1=23$, last term $a_n=1$, and common difference $d=-2$.
To find the number of terms, $a_n = a_1 + (n-1)d \Rightarrow 1 = 23 + (n-1)(-2) \Rightarrow -22 = -2(n-1) \Rightarrow 11 = n-1 \Rightarrow n=12$.
The sum of an arithmetic progression is $S_n = \frac{n}{2}(a_1+a_n)$.
$N = \frac{12}{2}(23+1) = 6(24) = 144$.
This matches option (3).

Answer

121

Answer

124

Answer

144

Answer

169

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