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How many four digit numbers can be formed using digits 0, 2, 3, 5, 7, 8. 

(a) How many of them are even 

(b) How many are divisible by 5 

(c) How many are greater than 5300 

(d) How many end with 7 

[Hint: Here digits can be repeated]

1 Answer

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Best answer

(a) Even numbers: 

The units place can be filled by 0,2 or 8 this can be done 3 ways 

∴ The number of ways is 5 × 6 × 6 × 3 = 540 ways.

(b) Number divisible by 5: 

The units place can be filled by 0 or 5 in 2 ways 

∴ the number of ways = 5 × 6 × 6 × 2 = 360 ways. 

(c) greater than 5300: 

Case 1: 5 is placed in thousands place. The 2nd place is done by (3 or 5 or 7 or 8) 4 ways 

∴ the number of ways = 1 × 4 × 6 × 6 = 144 ways. 

Case 2: If 7 or 8 is placed in thousandth place this can be done in 2 ways hundrenth place can be filled by any of the remaining six numbers 

∴ The number of ways = 2 × 6 × 6 × 6 = 432 ways. 

Hence the total number of ways is 432 + 144 = 576 ways.

(d) End with 7: Units place can be filled only one way the 1st place can be filled by 5 ways 

∴ the number of ways = 1 × 5 × 6 × 6 = 180 ways.

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