Latest Infosys Aptitude Question SOLUTION: how many five digit numbers are there such that two left most digits are even and remaining are odd and 4 digit should not be repeated.

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(#M40032761) INFOSYS Logical Reasoning question Keep an EYE

how many five digit numbers are there such that two left most digits are even and remaining are odd and 4 digit should not be repeated.

Asked In Infosys srijan g (10 years ago)

Solved Shankar Read Solution (15)

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Read Solution (15) :

srikanth  (10 years ago)

N=(4*5-1)*5*5*5=2375

where
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
1 case of 44 for two leftmost digit
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}

Is this solution Helpfull? Yes (33) | No (10)  

Shankar  (10 years ago)

even numbers are 2,4,6,8 and odd 1,3,5,7,9

so choice for first 2 digit 4 and 4

and for last 3 digit 5 ,4 ,3 coz not repeated

so ans 4*4*5*4*3 = 960

Is this solution Helpfull? Yes (32) | No (16)  

Mohammed Vaseem  (8 years ago)

Starting from the left most digit...

the possible combinations for first digit the possible ways are {2,4,6,8}=4 ways. (Here '0' is not included coz it won't make it a 5 digit number.)

Now for second digit total ways are (0,2,4,6,8) =5 ways.

but there is an exception that the digit '4' should not be repeated. so out of total cases i.e 4x5=20 , one such case exists and i.e. 44 remove it. so now the total ways will be 20-1 = 19 ways.

and for 3rd digit possible ways are {1,3,5,7,9} = 5 ways

similarly for 4th digit possible ways are {1,3,5,7,9} = 5 ways

similarly for 5th digit possible ways are {1,3,5,7,9} = 5 ways

hence the total number of ways are 19x5x5x5= 2375

Is this solution Helpfull? Yes (16) | No (1)  

abhik k. dey  (10 years ago)

for 1-st digit total case 2,4,6,8
for 2-nd digit total case 0,2,4,6,8
repeat nos are 22,44,66,88
for 3-rd, 4-th, 5-th digit total case 1,3,5,7,9 but not repeated

so total case (4*5-4)*5*4*3=960

Is this solution Helpfull? Yes (9) | No (2)  

Snehasis  (10 years ago)

4*4*5*4*5=1600

Is this solution Helpfull? Yes (8) | No (9)  

kunda anusha  (10 years ago)

can anybody giv the correct explanation

Is this solution Helpfull? Yes (3) | No (2)  

rahul singh  (10 years ago)

may be 960 numbers are there

Is this solution Helpfull? Yes (2) | No (4)  

veena mungara  (10 years ago)

pls give me the correct solution.

Is this solution Helpfull? Yes (2) | No (2)  

swastika chakraborty  (7 years ago)

even numbers are 0,2,4,6,8 and odd numbers are 1,3,5,7,9
so the left most digit can be any of {2,4,6,8) i,e. in 4 ways and second left digit can be any of {0,2,4,6,8} i.e. in 5 ways but as the case 44 is not permitted,hence two left digits can be chosen in (4*5)-1=19 ways
now remaining 3 numbers can be chosen in 5*5*5 ways(here repetition is allowed)
so, total case=19*125=2375

Is this solution Helpfull? Yes (2) | No   

vineetha  (8 years ago)

The first two slots need to be even numbers. The first one can be filled by 4 numbers (2, 4, 6, 8). The second can be filled by 4 numbers (0, 2, 6, 8) i.e., exclude 4 and include 0 in the second spot.

Hence, the first two can be filled up in 4 x 4 = 16 ways.

The remaining three slots have to be filled up using any of 1, 3, 5, 7, 9 digits as they can be repeated. i.e.,

third slot filled by 5 digits
fourth by 5
and fifth by 5

So, 5 x 5 x 5 = 125 ways of filling the last three slots.

Combining, we have 125 x 16 = 2000.

Is this solution Helpfull? Yes (1) | No (2)  

dilpreet  (10 years ago)

acc to me
even numbers are 2,4,6,8 and odd 1,3,5,7,9
1st 2 no even and next 3 odd
1st repeat 1 even then repeat 1 odd
so 4*4*5*4*3 + 4*3*5*4*4 = 1920

Is this solution Helpfull? Yes | No (2)  

pratham  (10 years ago)

even : 0,2,4,6,8
odd: 1,3,5,7,9

we cant take zero on left most size as this will not satisfy condition(5 digit num)
so we have 4 for first digit
5 for second left most
and 5 for 3rd digit
and we cant take digit selected on 3rd left most as it is given that 4th digit should nt repeat.
hence we have 4 choices for 4th digit and for final digit we can take 4 number excluding num selected on 4th place
hence ans should be
4*5*5*4*4 = 1600

Is this solution Helpfull? Yes | No (2)  

Ankit Goyal  (9 years ago)

if 4 is not used then 3*4*5*5*5 = 1500 and if 4 used then its 1*4*5*5*5 + 3*1*5*5*5 = 875 then total 1500 + 875 = 2375

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maharaja  (9 years ago)

samjh nahin aaya....????

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sudhanshu  (8 years ago)

4*4*5*5*5=2000.since at first position we can use only 2,4,6,8 not 0 but at second place we can use 0,2,6,8 not 4 because it should not be repeated . and at last 3 place we can arrange simply as 5*5*5 because there are 5 odd digit which can be repeated.

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