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Question
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Solution
Case I: Three-digit numbers with 4 occurring in hundred’s place:
100’s place digit can be selected in 1 way.
Ten’s place can be filled by anyone of the number 2, 3, 5, 6.
∴ 10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ Total number of numbers which have 4 in 100’s place = 1 × 4 × 3 = 12
Case II: Three digit numbers more than 500
100’s place digit can be selected in 2 ways.
10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ Total number of three digit numbers more than 500 = 2 × 4 × 3 = 24
Case III: Number of four-digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ Total four digit numbers formed = 5 × 4 × 3 × 2= 120
Case IV: Number of five digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ Total five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120
∴ Total number of numbers that exceed 400 = 12 + 24 + 120 + 120 = 276
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