M4maths Previous Puzzles

02 Sep 2011 number

How many numbers can be formed by using any 6 non zero digits. Repetition of digits is not allowed.

OPtion

1) 7*5*(3^4)*(2^6)
2) 7*(5^2)*(3^2)*(2^6)
3) (7^2)*5*3*2
4) 7*(5^2)*(3^3)*(2*6)
5) 7*5*(3^3)*(2^6)
6) (7^2)*(5^2)*(3^3)*2
7) 7*5*3*(2^6)
8) 7*5*3*2
9) (7^7)*5*3*2
10) none of these

Solution

answer is 9*8*7*6*5*4 = 7*5*(3^3)*(2^6)

Correct Option: 5 Best Solution (5)

Nishant    13 years AGO

since,we have to make six digit number without repitition of digits,so consider numbers from 1 to 9
and, _ _ _ _ _ _ from unit's place to lac place,
so,
for unit place,no. of ways=9
for ten's place,no. of ways=8
for hundred place,no. of ways=7
for thousand place,no. of ways=6
for ten thousand place,no. of ways=5
for lac place, no. of ways=4
so,total ways without repitition=9*8*7*6*5*4
=60480

Like? Yes (10) | No (2)  

Shourya Goyal   13 years AGO

since 9*8*7*6*5*4
= (3^2)*(2^3)*7*(2*3)*5*(2^2)
= 7*5*(3^3)*(2^6)

Like? Yes (2) | No (1)  

Himanshu Mittal   13 years AGO

from 9 digits we have to select 6 digits. 9!/(9-6)!=9*8*7*6*5*4=7*5*(3^3)*(2^6)

Like? Yes (1) | No (1)  

Arun kumar   13 years AGO

Max No of ways =
9_ 8_ 7_ 6 _ 5_ 4_
(Since it is a six digit no)
9*8*7*5*4 = 60480

Like? Yes (1) | No (1)  

Perumal Natarajan   13 years AGO

From 9 digits(non Zero) we can take any 6 digit numbers. so the combination is 9!/(9-6)!=9*8*7*6*5*4
=(3^2)*(2^3)*7*(2*3)*5*(2^2)
=7*5*(3^3)*(2^6)

Like? Yes (1) | No (1)