Question

How many different numbers, greater than 5000 can be
formed with the digits 1, 2,5,9,0 when repetition of digits is
not allowed ?
​

Answer 1

Answer:

To make a four- digit number greater than 5000 , we have 3 choices (5,6 ,7) for our thousands place.

Similarly, for our hundreds place we have 4 choices (3,4, 5/6/7, 5/6/7) and then 3, and 2 .

So, the total number of possibilities will be

3×4×3×2 = 7 2

These will be: (just for insight)

{ 5347,5376,5437,6734,5743,.....7346,7354,7435,7453,7534,7643 ... }

Also, all 5 - digit numbers formed from these digits will be greater than 5000 . So it will be 5P5 = 5! = 120

Summing up, the total number of natural numbers formed from the digits 3,4,5,6,7 without repetition, and which are greater than 5000 are 192 .

Now, for the even case, we need to have an even number (here, 4,6 ) at the ones place.

Therefore, we will consider two cases each for 4 and 5 - digit numbers.

Case 1: 4 has pre- occupied the ones place in the 4 -digit number.

Now we have to arrange 4 numbers into 3 places such that the complete thing is still greater than 5000 .

Thus, 3 choices (5,6,7) for the thousands place, 3(3,5/6/7,5/6/7) for the hundreds and 2(3/5/6/7,3/5/6/7) for the tens, giving us 18 numbers.

Case 2: 6 has pre- occupied the ones place in the 4 -digit number.

This time we will have 2×3×2 = 12 numbers.

Now, by similar arguments for the 5 - digit numbers, we get a total of 2×4! = 48 numbers.

We thus end up with a total of 78 even numbers and hence also 114 odd numbers.

Hope that helped.