@invAMARISH said: total no of song=14, Total no of way 2^12*5c1*6c1=2^12*30

@chirag1501 said: u can write it as, n*(n+1)*(n-1)*(5n+2) (n+1)*(n-1) means whatever value of n you take, you will always have one odd and one even number in the expression, hence it will always be divisible by 6

there are 11 cases when sum of 5 numbers is 17, and out of these 11 cases 9 times 1 number out of 5 numbers is 0 and a can not be 0

so when a+c+e+g+i = 17, and one of them is 0 => 4*4! ways
and b+d+f+h+j = 28 => 5! ways
so total 4*4!*5! ways,
there are 9 such cases, so 9*4*4!*5!

when a+c+e+g+i = 17, (when none of the 0), there are 2 such cases , so 2*5!
and b+d+f+h+j = 28 => 2*5! ways
total 2*5!*5!

same for a+c+e+g+i = 28, and b+d+f+h+l = 17 => 9*5!*5! + 2*4*4!*5! ways

abcdefghij is a 10 digit number

a*10^9 + b*10^8 + c*10^7 + d*10^6 + e*10^5 + f*10^4 + g*10^3 + h*10^2 + i*10 + j mod 11 = 0

-a + b - c + d - e + f - g + h - i + j mod 11 = 0

(b+d+f+h+j) - (a+c+e+g+i) mod 11 = 0

a+b+c+d+e+f+g+h+i+j = 45

so we should divide 45 in 2 parts such that their difference is divisible by 11

only 28 and 17 is possible

cases when sum of 5 distinct single digit numbers is 17

0 1 2 5 9
0 1 2 6 8
0 1 3 4 9
0 1 3 5 8
0 1 3 6 7
0 1 4 5 7
0 2 3 4 8
0 2 3 5 7
0 2 4 5 6
1 2 3 4 7
1 2 3 5 7

11 cases

if a+c+e+g+i = 17 and b+d+f+h+j = 28 =>9*4*4!*5! + 2*5!*5! ways (as a can not be 0)

if a+c+e+g+i = 28 and b+d+f+h+j = 17 => 2*4*4!*5! + 9*5!*5! ways

total = 9*4*4!*5! + 2*5!*5! + 2*4*4! + 9*5!*5!

=> 11*4*4!*5! + 11*5!*5! = 285120 numbers

Own->belonging to person,personal, peculiar
profane->not religious,secular
parches->earth cracks due to intense heat
totemic->practice of having animal as symbol of family,group,clan
implacable->unstoppable
defunct->dead,stop
atone->make amends by paying compensation.In war team who has defeated other pays fine for helping
buttressed->support
reprobate-> villain,scoundrel
beset->surrounded by
surmount->conjecture

1->secular
2->snuff's out
3->risen
4->implacable
5->atoned
6->hope