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
  
  
  
• 11 years agoHelpfull: Yes(12) No(2)
• 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
• 11 years agoHelpfull: Yes(4) No(2)
• 120 ways.....
  let 2,3,4,5,6 in each dice respectively....it can permutate in 5! ways
  hence 120 ways
  
• 11 years agoHelpfull: Yes(3) No(47)
• I think its 7741 times shows greater than 15.
• 11 years agoHelpfull: Yes(3) No(5)
• 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
• 11 years agoHelpfull: Yes(3) No(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
  
• 10 years agoHelpfull: Yes(3) No(0)
• Abhishek - can u plz explain your ans?
• 11 years agoHelpfull: Yes(2) No(1)
• 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
• 11 years agoHelpfull: Yes(2) No(2)
• 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
• 11 years agoHelpfull: Yes(2) No(4)
• remainder=4
  in the numarator we get last digit 0
  in the denominator last digit we get is 6
  hence remainder is 4
  
• 10 years agoHelpfull: Yes(1) No(0)
• 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
  
• 10 years agoHelpfull: Yes(1) No(0)
• plz explain any one clearly
• 10 years agoHelpfull: Yes(1) No(0)
• Can u plz explain it?
• 11 years agoHelpfull: Yes(0) No(2)
• giveme solutions also
• 10 years agoHelpfull: Yes(0) No(0)