# The number of 6 - digit number s that can be made with the digits 0,1,2,3,4&5 so that even digits occupy odd places is ( A ) 24 (B) 36 ( C ) 48 ( D ) NONE

Vikas TU
14149 Points
7 years ago
_   _   _   _   _   _
1    2   3   4   5  6     => Positions
Now here 1 3 and 5 would occupy only even numbers and that is=> 0,2 and 4 with 0 cannot cover in first positon which would discard the six digit formation number.
Therfore at first positon no of ways = > 2
at 3rd and 5th postion no of ways => 3
at at other evn places any number could occupy.
therfore no of ways would be = > 6 for all three.
hence total ways => for repeating
=> 6*6*6*3*3*2

T Jahnavi
40 Points
7 years ago
Ankit Jaiswal
165 Points
7 years ago
the numbers which are allowed are 0,1,2,3,4,5
and the number has to be 6 digit ( _ _ _ _ _ _ ) so 0 cannot occupy lacks place (the number should not start with 0 as 0_ _ _ _ _ ) as it will be 5digit
so the remaining case are 4
2_0_4_
2_4_0_
4_0_2_
4_2_0_
now there are 4 cases with total 3 places remaining with 3 different numbers (1,3,5)
so there are total 3! ways for them to occupy the 3 places
and hence totol number of 6 digit numbers possible are
6+6+6+6=6*4=24

Ankit Jaiswal
165 Points
7 years ago
i suppose i dont need to metion that 3!=3*2*1=6
Siddharth
19 Points
5 years ago
Three odd places occupy by three even number.so total no. Of arrangements=3!We don`t need 0 at first place so no. Of arrangements keeping 0 at first place is 2!Total rearrangement excluding 0 is( 3!-2!)For even places 3!Total re arrangement( 3!-2!)×3!=24