Iift 2010 Question 2
This is a detailed step-by-step solution for Iift 2010 Question 2. Understand the concept, common mistakes, and expert tips to solve similar questions in IIFT exam.
IIFT 2010
We are given exactly 7 digits - 3, 4, 4, 5, 6, 6, 7. The millions digit can be either 6 or 7.
Case 1: When the millions digit is 6.
6 _ _ _ _ _ _ We are left with six digits {3, 4, 4, 5, 6, 7}.
These six digits can be arranged in six places in 6!/2! ways.
Case 1: When the millions digit is 7.
7 _ _ _ _ _ _ We are left with six digits {3, 4, 4, 5, 6, 6}.
These six digits can be arranged in six places in 6!/2! x 2! ways.
Therefore, total number of numbers 'n' = 6!/2! + 6!/2! x 2! = 360 + 180 = 540. Hence, option C is the correct answer.