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JEE Main 20264 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Permutation and Combination

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Permutation and Combination

Let A=\ (a,b,c): a,b,c are non-negative integers and a+b+2c=22\ ....

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

Options

  1. A121121
  2. B124124
  3. C144144
  4. D169169

Correct answer

C. 144144

Step-by-step solution

The given equation is a+b+2c=22a + b + 2c = 22, where a,b,ca, b, c are non-negative integers.

This can be rewritten as a+b=22−2ca + b = 22 - 2c.

Since a≥0a \ge 0 and b≥0b \ge 0, we must have 22−2c≥022 - 2c \ge 0, which gives c≤11c \le 11.

Since cc is a non-negative integer, the possible values for cc are 0,1,2,…,110, 1, 2, \dots, 11.

For a fixed value of cc, the number of non-negative integer solutions to the equation a+b=22−2ca + b = 22 - 2c is given by (22−2c)+1=23−2c(22 - 2c) + 1 = 23 - 2c.

The total number of solutions n(A)n(A) is the sum of the number of solutions for each possible value of cc:

n(A)=∑c=011(23−2c)n(A) = \sum_{c=0}^{11} (23 - 2c)

n(A)=23+21+19+⋯+1n(A) = 23 + 21 + 19 + \dots + 1

This is an arithmetic progression with 1212 terms, first term 11, and last term 2323.

n(A)=122(1+23)=6×24=144n(A) = \dfrac{12}{2} (1 + 23) = 6 \times 24 = 144

Answer: 144144
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