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