Latest TCS Aptitude Question SOLUTION: How many 6-digit even numbers can be formed from the digits 1,2,3,4,5,6 and 7 so that digits should not repeat and the second last digit is even? a)720 b)320 c)2160 d)6480

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Permutation and Combination

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40023675) TCS Numerical Ability question Keep an EYE

How many 6-digit even numbers can be formed from the digits 1,2,3,4,5,6 and 7 so that digits should not repeat and the second last digit is even?
a)720
b)320
c)2160
d)6480

Asked In TCS Dipannita Adhikary (10 years ago)

Unsolved Read Solution (9)

Is this Puzzle helpful?   (33)   (12) Submit Your Solution Permutation and Combination

Read Solution (9) :

Karthika  (10 years ago)

EDIT in the above answer**
------
5P4 * 6 since there are 6 cases

5P4*6=720

Is this solution Helpfull? Yes (24) | No (9)  

TCS 2014  (10 years ago)

ans is 6*5*4*3*3*2=2160
the second last digit is 2,4,6 only, so there is 3 posibilites to fill up the 3rd place..we have total 7 num, so 6th place can be fill with 6number, 5th place fill with 5 number and so on..

Is this solution Helpfull? Yes (16) | No (4)  

Karthika  (10 years ago)

I hope this is the answer!
To satisfy the above conditions as in the question..(the second last and last number shuld be even)
6 cases
_ _ _ _ 42
_ _ _ _ 62
_ _ _ _ 24
_ _ _ _ 64
_ _ _ _ 26
_ _ _ _ 46

So 2 digits are already set as above
Now take the first example. _ _ _ _ 4 2
here the remaining digits that can be filled are 1,3,5,6,7
So 5P3

Since 6 such cases are there 5P3*6=720

Is this solution Helpfull? Yes (11) | No (13)  

Hemanta Nandi  (10 years ago)

As the second last digit is even and the number is even, So that we have to fill up the last two digits with even numbers i.e. 3P2 .
And the remaining four positions we can fill up in 5*4*3*2 ways.
So that our required answer is 3P2*5*4*3*2 = 6*5*4*3*2 = 720

Is this solution Helpfull? Yes (9) | No (5)  

Dheeraz  (10 years ago)

positions 1 2 3 4 5 6

5th should be even so posibility is 2,4,6=3
other 5 can be any among 6 6p5=6!
ans=6!*3=2160

Is this solution Helpfull? Yes (5) | No (2)  

Ashish kalra  (10 years ago)

ans is 720 expl: for unit digit we have three options 2,4,6 then for second last 2 options left as one out of 2,4,6 used in unit digit then for remaining 5 places 5*4*3*2*1 so the ans is 5*4*3*2*2*3=720

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

sumana  (10 years ago)

3c2*2!*5*4*3*2=720

Is this solution Helpfull? Yes | No (3)  

prajval  (7 years ago)

5!*3*2=720

Is this solution Helpfull? Yes | No (1)  

SANTOSH MOHANTY  (7 years ago)

So according to question.......... making 6 digit even number.
....
..
.
.
.
2nd last digit is even( i.e 5th position must even)

(1th)_____ (2nd) ______ (3rd) ______ (4th) _______ (5th) Even (6th)___________-
.
.
.we have 3 even number........... Now number of way selecting 3 even number for one place( 5th position) = 3C1=3
.
.
Remaining place( 1st, 2nd, 3rd, 4th, 6th)
Total number remaining is six because out of 7 number one must be fixed at 5th place( only even number)
..
.
.
number of arranging 6 number in 5 vacant place is = 6P5 = 6*5*4*3*2*1
.
.
Now total number of digit form is = 6P5*3=2160
....
AS per position (1th)6 (2nd)5 (3rd)4 (4th)3 (5th)( 3c1=3) (6th)2 (7th)1
== 6*5*4*3*3*2*1= 2160

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.