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The no. Of six digit number that can be formed from tje digits 0,1,2,3,4,5,6&7 so that digits do not repeat and the terminal digits are even

The no. Of six digit number that can be formed from tje digits 0,1,2,3,4,5,6&7 so that digits do not repeat and the terminal digits are even

Grade:11

7 Answers

Ankit Jaiswal
165 Points
8 years ago
the total number form which the 6 digit number should be made are 8
out of which 4 are even (0,2,4,6) and these are to be arranged in the terminal points (end points)
hand hence we have to arrange 4 numbers in 2 places (even             even)
hence total numbers of ways we can do this are 4!/(4-2)! = 4*3 = 12
now we have 12 cases with terminal points occupied with two numbers 
so we have 4 places remaining with 6 numbers remaining and the number of ways we can do this are 
6!/(6-4)! = 6*5*4*3 = 360  
hence we have 360 solutions in each case and there are total 12 cases 
therefore the total number of 6-digit number that satisfies our condition =
360*12 = 4320
Ankit Jaiswal
165 Points
8 years ago
please approve the and
T Jahnavi
40 Points
8 years ago
Sir but the answer is 720.......................................................................................................................
Ankit Jaiswal
165 Points
8 years ago
 OOH I am sorry for a mistake
Actually I considered that all numbers can occupy first position ( THIS                ) but it cannnot be occupied by 0 
so only the number of cases will be affected 
that is the first place will have 3 possible cases(2,4,6) and last place will aslo have 3 possible cases(0,2,4,6 out of which oneis already in first place)
so our cases reduces to 3*3 = 9 instead of  12
but then also the answer will be 360*9 =3240
 
 
and please don’t call me sir.
 
Ankit Jaiswal
165 Points
8 years ago
i think that the answer will be 720
Ankit Jaiswal
165 Points
8 years ago
i think that the answer will be 720
only is “0” is not incuded
 
And for that question the solution will be
let the number be 
a b c d e f (say)
a = 3 numbers possible (2,4,6)
f = 2 numbers possible (2,4,6 out of which one has occupied “a”)
and then it goes on decreasing
 
b = 5 numers possible
c = 4 numbers possible
d = 3 numbers possible
e = 2 numbers possible
and hence the answer should be 
3*2*5*4*3*2 =720
 
 
previous 
ARKADEEP DAS
44 Points
7 years ago
the answer is simple parmutation
8p3 =  56
so its done
its easy and simple
do u got it it’s nice and simple
 

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