Latest TCS Aptitude Question SOLUTION: In how many ways we can get a sum greater than 15 by throwing 5 distinct dice.

651 ways
Explanation:
Coefficient of x^15 in (x+x^2+x^3+x^4+x^5+x^6)^5
i.e coefficient of x^10 in (1+x+x^2+x^3+x^4+x^5)^5
means coefficient of x^10 in ((1-x^6)/(1-x))^5
coefficient of x^10 in (1-x^6)^5(1-x)^(-5)
ie in (1-5C1x^6)(1-x)^(-5)

=14C10-(5C1)*(8C4)
=1001-5*70=651

Coefficient of x^15 in (x+x^2+x^3+x^4+x^5+x^6)^5
i.e coefficient of x^10 in (1+x+x^2+x^3+x^4+x^5)^5
means coefficient of x^10 in ((1-x^6)/(1-x))^5
coefficient of x^10 in (1-x^6)^5(1-x)^(-5)
ie in (1-5C1x^6)(1-x)^(-5)

=14C10-(5C1)*(8C4)
=1001-5*70=651

120 ways.....
let 2,3,4,5,6 in each dice respectively....it can permutate in 5! ways
hence 120 ways

I think its 7741 times shows greater than 15.

this the explanation of above solution given by me
if u want to calculate the no of ways to get 15 by throwing 5 dices ...u may compare the same case wid :
solution of
a+b+c+d+e=15
where 1

7658 ways.


Case 1.(All five position are fixed and have same number say x )

possible values of x are 6,5,4.
total ways = 3.


Case 2.(Four out of 5 position are fixed and have same number say x)

when x=6 then number of ways = 6*5C1=30
when x=5 then number of ways = 6*5C1=30
when x=4 then number of ways = 6*5C1=30
when x=3 then number of ways = 3*5C1=15
values of 2 and 1 for x are not possible in this case.
therefore total number of ways = 105



Case 3.(3 out of 5 position are fixed and have same number say x)

when x=6 then number of ways = (6*6)*5C2=360
when x=5 then number of ways = (6*6)*5C2=360
when x=4 then number of ways = (4*6)*5C2=240
when x=3 then number of ways = (1*6)*5C2=60
when x=2 then number of ways = (1*3)*5C2=30
value of 1 is not possible in this case.
therefore total number of ways = 1050


Case 4.(2 out of five position are fixed and have same value say x)

when x=6 then number of ways = (5*6*6)*5C3=1800
when x=5 then number of ways = (3*6*6)*5C3=1080
when x=4 then number of ways = (1*6*6)*5C3=360
when x=3 then number of ways = (1*4*6)*5C3=240
when x=2 then number of ways = (1*2*6)*5C3=120
when x=1 then number of ways = (1*1*5)*5C3=50

therefore total number of ways = 3650



Case 5.(1 out of 5 position are fixed and have number say x)

when x=6 then number of ways = (1*5*6*6)*5C4=900
when x=5 then number of ways = (1*4*6*6)*5C4=720
when x=4 then number of ways = (1*3*6*6)*5C4=540
when x=3 then number of ways = (1*2*6*6)*5C4=360
when x=2 then number of ways = (1*1*6*6)*5C4=180
when x=1 then number of ways = (1*1*5*6)*5C4=150

therefore total number of ways = 2850

Now sum total of number of ways of all the cases = 3+105+1050+3650+2850 = 7658

Abhishek - can u plz explain your ans?

if u want to calculate the no of ways to get 15 by throwing 5 dices ...u may compare the same case wid :
solution of
a+b+c+d+e=15
where 1

first prob is 1,2,3,4,5=15 we arrange in 5!=120 ways
second prob is 6,4,1,2,3=15 it also in 5!=120 ways

other porb can't distinct

So total =120+120=240

remainder=4
in the numarator we get last digit 0
in the denominator last digit we get is 6
hence remainder is 4

There are 15 ways as follows
1. 4+2+3+6+1=16
2. 5+2+3+6+1=17
3. 6+2+3+6+1=18
4. 4+3+3+6+1=17
5. 4+4+3+6+1=18
6. 4+5+3+6+1=19
7. 4+6+3+6+1=20
8. 4+2+4+6+1=17
9. 4+2+5+6+1=18
10. 4+2+6+6+1=19
11. 4+2+3+6+2=17
12. 4+2+3+6+3=18
13. 4+2+3+6+4=19
14. 4+2+3+6+5=20
15. 4+2+3+6+6=21

plz explain any one clearly

Can u plz explain it?

giveme solutions also