Bunuel wrote:
eresh wrote:
How many possible numbers can be built from the set 3,5,7,8 and 9 so that the number is greater than 7000?
I assume the condition is that the numbers cannot repeat.
If so then:
If we have 4 digit number: first digit must be 7, 8 or 9.
If we have 5 digit number any number can be the first digit.
4 digit number:
3C1*4C3*3!=72 (3C1 to choose 1 # from 7,8 or 9; 4C3 to choose the rest 3 numbers for 4 digit number out of 4 left; 3! # of combinations of these 3 numbers)
5 digit number: 5!=120
72+120=192
Hi
Bunuel,
Just wondering, I assumed no repetition, similar to your assumption. While i understand your method....just wanted to know if there is wrong with this method
No of 4 digit number - 3C1(for 7,8,9)* 4C1(since 7/8/9 cannot be repeated, 1 number chosen, 4 left to pick one)*3C1(similarly 2 numbers are chosen already,3 left)*2C1(3 numbers are chosen already, 2 left)
Hope my solution is clear. Appreciate your help.