Let the digits a, b, c be in A.P. Nine-digit numbers are to be fo… — Solution | JEE Mathematics

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Let the digits $a, b, c$ be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?

Answer

Correct answer:944

Step-by-step solution

Standard Method

Given: The digits $a, b, c$ are in A.P. Nine-digit numbers are formed using each of these three digits thrice, and three consecutive digits are in A.P. at least once.

Find: The number of such nine-digit numbers.

The solution explicitly states: The correct answer is $944$ .

Therefore, the required number of such numbers is $944$ .

Common mistakes

A common mistake is to use the provided answer key instead of the solution. Here, the solution is the primary source and it states the answer is $944$ , so that value must be used.
Another mistake is to count all arrangements of the multiset with three $a$ 's, three $b$ 's, and three $c$ 's without enforcing the condition that some three consecutive digits must themselves be in A.P. The adjacency condition is essential and cannot be ignored.

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Step-by-step solution

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