JEE Mains · Maths · STD 11 - 6. permutation and combination
The number of ways of selecting two numbers \(a\) and \(b\), \(a \in\{2,4,6, \ldots ., 100\} \quad\) and \(\quad b \in\{1,3,5, \ldots ., 99\}\) such that \(2\) is the remainder when \(a+b\) is divided by \(23\) is\(.......\)
- A \(109\)
- B \(110\)
- C \(108\)
- D \(154\)
Answer & Solution
Correct Answer
(C) \(108\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}a \in\{2,4,6,8,10, \ldots ., 100\} \\b \in\{1,3,5,7,9, \ldots ., 99\}\end{array}\) Now, \(a + b \in\{25,71,117,163\}\) \((i)\) \(a+b=25\), no. of ordered pairs \((a, b)\) is \(12\) \((ii)\) \(a+b=71\), no. of ordered pairs \((a, b)\) is \(35\) \((iii)\)…
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