4 cannot appear more than once

sanju09 wrote:How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?
(A) 1875
(B) 2000
(C) 2375
(D) 2500
(E) 3875


[spoiler]Source: Eric’s collection on BTG


OA C[/spoiler]

Here following cases are possible:-
1) when 4 is present at the leftmost digit.
2) when 4 is present at the digit adjacent to the left most digit,
3) when 4 is not present at any of the leftmost places;

case 1) -,-,-,-,-; when 4 is present at the left most digit, then the digit adjacent to it can be filled (by any of the (0,2,6,8)) in 4 ways, and the remaining places can be filled by the 5 odd numbers in 5*5*5 ways; hence total no. of ways= 1*4*5*5*5;

case 2)-,-,-,-,-; when 4 is not there at the left most digit, it can be filled ( by any of the 2,6,8) in 3 ways, and the digit adjacent to the leftmost digit will be filled by the 4 in 1 way; and the remaining places can be filled by the 5 odd numbers in 5*5*5 ways; hence total no. of ways= 3*1*5*5*5;

case 3)-,-,-,-,-; when 4 is not selected for the arrangement; then the left most digit can be filled ( by any of the 2,6,8) in 3 ways, and the digit adjacent to leftmost digit can be filled ( by any of the 0,2,6,8) in 4 ways, and the remaining places can be filled by the 5 odd numbers in 5*5*5 ways; hence total no. of ways = 3*4*5*5*5;
hence total required no. of ways= 1*4*5*5*5+3*1*5*5*5+3*4*5*5*5=2375 hence C

sanju09 wrote:How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?
(A) 1875
(B) 2000
(C) 2375
(D) 2500
(E) 3875


[spoiler]Source: Eric’s collection on BTG


OA C[/spoiler]

Hi,
1st digit can be chosen from 2,4,6,8 -> 4ways
2nd digit can be chosen from 0,2,4,6,8 -> 5 ways
3rd,4th,5th digits can be chosen from 1,3,5,7,9 -> 5.5.5 = 125 ways
So, number of ways = 4.5.5.5.5 = 2500

We need to remove the case of 1st two digits being 4. No. of ways of doing this is 1.1.5.5.5 = 125.

So, the required number of ways is 2500-125 = 2375
Hence, answer C .

Cheers!

Hi Frank

Is there any other option to get the result directly without subtracting the undesired part? I mean solving with just one transaction.

Thnx

Ozlemg wrote:Hi Frank

Is there any other option to get the result directly without subtracting the undesired part? I mean solving with just one transaction.

Thnx

Hi,
You have to consider 2 cases even in this method.
Case 1:1st digit is not 4.
So, 1st digit is chosen from {2,6,8} in 3 ways.
Each of the remaining 4 digits can be chosen in 5 ways.
So, number of 5 digit numbers = 3.5^4=1875.

Case 2:1st digit is 4
so, 2nd digit can be chosen from {0,2,6,8} in 4 ways
Each of the remaining 3 digits can be chosen in 5 ways.
So, number of 5 digit numbers = 4.5^3=500.

So, total number of 5 digit numbers = 1875+500=2375.
(or)
If you are looking for one transaction, you can follow this approach, which is effectively same:
First 2 digits can be chosen in 4C1.5C1-1 =19 ways. We subtracted 1 because we need to subtract the case of '44'.
Each of the remaining 3 digits can be chosen in 5 ways.
So, number of 5 digit numbers = 19.5^3=2375.

I am not sure if this is what you are looking for, because it is almost same as the one I posted earlier.

5 digit no. -----
eeooo
3rd,4th,5th numbers are odd and thus no constraint on them as such
options for 3rd=5 (1,3,5,7,9)
options for 4th=5 (1,3,5,7,9)
options for 5th=5 (1,3,5,7,9)
1st digit can't be 0. ( if it is, then it becomes 4 digit no.)
for 1st digit, if 4 selected - option =1 (we selected 4)
then for 2nd digit, we can't select 4 and thus options =4 (0,2,6,8)
no.=1*4=4
if 4 not selected for 1st position, then options for 1st = 3 (2,6,8)
options for 2nd=5 (0,2,4,6,8)
no.=3*5=15
options for 1st and 2nd combines=4+15=19
thus combining with 3rd,4t and 5th position = 19*5*5*5 = 2375
IMO C