# How many numbers greater than 1200 can be formed with the digits 3, 4, 5, 6, 7 and 8 if a digit cannot occur more than once in a number?

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### 2 Answers

We have the digits 3, 4, 5, 6, 7 and 8 and we need to create numbers greater than 1200. Now, any number formed with four of the digits is greater than 1200. Also any number containing all the 5 digits is greater than 1200.

We have to find the number of permutations when 4 out of 5 digits are chosen. This is given by the expression 5! / (5 – 4)!

= 5! / 1!

= 120

Also, the number of permutations of choosing all 5 digits is 5! / (5 – 5)! = 5! / 1 = 120.

Adding the two we get a total of 240 permutations.

**Therefore 240 different numbers can be created with the digits 3, 4, 5, 6, 7 and 8 which are greater than 1200.**

To find the number of numbers greater than 1200 that can be formed with 6 digits 3, 4,5,6 and 8 without repetition:

We see that all the digits 3,4,5, 6 and 8 are greater than the digits 1 and 2.

Therefore any 4 digit numbers formed with the given digits are greater than 1200. Also any 5 and 6 digit numbers formed are geater than 1200. Also there is a condition of no repetetion.

The number of 4 digit numbers with the given 6 digits without repetion is 6P4 = 6*5*4*3 = 360.

The the number of 5 digit numbers that can be formed without repetition with 6 given digits = 6*5*4*3*2 = 720 ways.

The number ways the 6 digit numbers using all the 6 digits without repetitions = 6! = 6*5*4*3*2*1 = 4320.

Therefore the number of numbers formed using the given six digits without repetition is 6P4+6P5 +6P6 = 360+720+ 720 = 1800.