Latest TCS Aptitude Question SOLUTION: How many six digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the numbers should not repeat and the second last digit is even?

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How many six digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that the numbers should not repeat and the second last digit is even?

Asked In TCS Laado Vadher (10 years ago)

Solved venkata satya kumar Read Solution (15)

Is this Puzzle helpful?   (9)   (3) Submit Your Solution Permutation and Combination

Read Solution (15) :

venkata satya kumar  (10 years ago)

here we have 7 numbers.we have 3 even numbers so that last digit can be arranged in 3 ways.second last digit can be arranged in 2 ways.the other digits can be arranged in 5,4,3,2 ways so the final answer is 3*2*5*4*3*2=720 ways

Is this solution Helpfull? Yes (47) | No (4)  

Sourav Mohanty  (10 years ago)

6*5*4*3*3*2
second last disgit has 3 possibilities,so remaining 6 no.s can be placed in other vacancies.

Is this solution Helpfull? Yes (13) | No (1)  

sai  (10 years ago)

last digit can be filled in 3 ways...2,4,6
second last digit cn be filled in 2ways...
remaining 4 digits can be filled in 5p4
total nubers= 5p4*2*3

Is this solution Helpfull? Yes (6) | No   

madhuri  (10 years ago)

6!=720 ways

Is this solution Helpfull? Yes (3) | No (7)  

NAMRATA KASHYAP  (9 years ago)

it will be 720 as it is solved in 5*4*3*2*2*3= 720

Is this solution Helpfull? Yes (3) | No   

sumeet kumar  (10 years ago)

suppose we have 6 blocks say.. _ _ _ _ _ _
now according to the question last block will be even and second last block will also be even...so for 2nd block element out of three that is...2,4,6 one can be selected in 3c1 ways and for last block in 2c1 ways...and rest can be arranged in 5! ways

and by multiplication rule...5!*3c1*2c1=720 ways is the required answer

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

DHIRENDRA   (10 years ago)

six digit even no means unit place must be even i.e.at unit place must be 1,3,5,7

6c5*4!=6!/(5!*1!)*4!
6*4!
6*4*3*2*1=144

answer=144

Is this solution Helpfull? Yes (1) | No (10)  

siva  (10 years ago)

720.
(5*4*3*2)*6

Is this solution Helpfull? Yes (1) | No   

hemant panchal  (10 years ago)

no. of odd digits = 4 = (1,3,5,7)
no. of even digits = 3 = (2,4,6)
no. should be 6 digit no
so,
(1st) (2nd) (3rd) (4th) (5th) (6th)
1357(246)
ways to arrange 3 even digits at the last 2 places = 3 P 2 = 3! = 6
ways to arrange 4 odd digits at 1st 4 places = 4 P 4 = 4! = 24
so,
no. of ways of arranging such 6 digits = 6*24 = 144

Is this solution Helpfull? Yes (1) | No (5)  

satyam  (9 years ago)

2*3*4*5*2*3

Is this solution Helpfull? Yes (1) | No   

madhuri  (10 years ago)

6!=720 numbers can be formed

Is this solution Helpfull? Yes | No (7)  

rohith kumar  (10 years ago)

here we have 3 even numbers so they are arranged in 3! ways, similarly the remaining 4 numbers are arranged in 4! ways so the answer is 3!*4!=144

Is this solution Helpfull? Yes | No (7)  

jagadeesh kumar.B  (10 years ago)

sinc the 2nd last digit is even, we have to arrange 2,4,6 in that place as 3p1 ways and the remaining 5 places are arranged with the remaing digits(6 digits) as 5p6, hence the required ans is 3p1+5p6= 3 ways...

Is this solution Helpfull? Yes | No (6)  

harsha vardhan veeturi  (10 years ago)

it should be 144 because we can fill the last digit with only 3 cases and last but one by 2 and remaining 4 places 4 digits ie 4! so finally solution is
4!*2*3=144

Is this solution Helpfull? Yes | No (4)  

subrahmanyam  (10 years ago)

selecting the appropriate digits is
3*4*3*2*1*2=144

Is this solution Helpfull? Yes | No   

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