2 votes 2 votes papesh asked Nov 5, 2016 papesh 1.2k views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply papesh commented Nov 5, 2016 i edited by papesh Nov 5, 2016 reply Follow Share i'm getting for 20 = 51271 for 21= 4920 plz check it! 0 votes 0 votes Digvijaysingh Gautam commented Nov 5, 2016 reply Follow Share I am getting 47085 for 20th question 0 votes 0 votes papesh commented Nov 5, 2016 reply Follow Share Previously I was also got 47085 ....plz check once again 0 votes 0 votes Digvijaysingh Gautam commented Nov 5, 2016 reply Follow Share 4920 for 21st ? 0 votes 0 votes papesh commented Nov 5, 2016 reply Follow Share @ Digvijaysingh Gautam can u explain with complete solution 0 votes 0 votes Digvijaysingh Gautam commented Nov 5, 2016 reply Follow Share is it correct ? 0 votes 0 votes papesh commented Nov 5, 2016 i edited by papesh Nov 5, 2016 reply Follow Share for 21 i think it's right i have got the same... i don't have ans.. 3c1 [103c3] -3c2 [92c3] =4920 0 votes 0 votes Digvijaysingh Gautam commented Nov 5, 2016 reply Follow Share 21) 3 cases possible case i) x1 > 40 so smallest value of x1 is 41 x1 + 41 + x2 + x3 = 100 x1 + x2 + x3 = 59 so possible combinations = 59+3-1C59 = 61C59 = 1830 case ii) x2 > 40 and x1<=40 first we will find number of combination such that x2 > 40 and then subtract the combinations of x2 > 40 and x1 > 40 first we will find number of combination such that x2 > 40 x1 + x2 + 41 + x3 = 100 x1 + x2 + x3 = 59 so possible combinations = 59+3-1C59 = 61C59 = 1830 Now number of combination such that x2 > 40 and x1 > 40 x1 + 41 + x2 + 41 + x3 = 100 x1 + x2 + x3 = 18 so possible combinations = 18+3-1C18 = 20C18 = 190 Hence number of combination such that x2 > 40 and x1<=40 = 1830 - 190 = 1640 case iii) x3 > 40 and x1 <= 40 and x2 <= 40 first we will find number of combination such that x3 > 40 and then subtract the combinations of x3 > 40 and x1 > 40 and also subtract the combinations of x3 > 40 and x2 > 40 first we will find number of combination such that x3 > 40 x1 + x2 + x3 + 41 = 100 x1 + x2 + x3 = 59 so possible combinations = 59+3-1C59 = 61C59 = 1830 Now number of combination such that x3 > 40 and x1 > 40 x1 + 41 + x2 + x3 + 41= 100 x1 + x2 + x3 = 18 so possible combinations = 18+3-1C18 = 20C18 = 190 Now number of combination such that x3 > 40 and x2 > 40 x1 + x2 + 41 + x3 + 41= 100 x1 + x2 + x3 = 18 so possible combinations = 18+3-1C18 = 20C18 = 190 Hence number of combination such that x3 > 40 and x1 <= 40 and x2 <= 40 = 1830 - 190 - 190 = 1450 So total ways = 1830 + 1640 + 1450 = 4920 2 votes 2 votes Digvijaysingh Gautam commented Nov 5, 2016 reply Follow Share @Gabbar see this answer to 20th question x1 + x2 + x3 + x4 =100 now x1 can take value from 0 to 10. So we will substitute the values one by one and find number of combinations 1) 0 + x2 + x3 + x4 = 100 => x2 + x3 + x4 = 100 => 100+3-1C100 = 102C100 = $\frac{101*102}{2}$ 2) 1 + x2 + x3 + x4 = 100 => x2 + x3 + x4 = 99 => 99+3-1C99 = 101C99 = $\frac{100*101}{2}$ 3) 2 + x2 + x3 + x4 = 100 => x2 + x3 + x4 = 98 => 98+3-1C98 = 100C98 = $\frac{99*100}{2}$ . . and so on 11) 10 + x2 + x3 + x4 = 100 => x2 + x3 + x4 = 90 => 90+3-1C90 = 92C90 = $\frac{91*92}{2}$ now we will have to add all these total combination = summation of n(n+1)/2 where n from 91 to 101 = summation of $\frac{n^{2}+n}{2}$ = $\frac{n(n+1)(2n+1)}{6*2}$ + $\frac{n(n+1)}{2*2}$ first put n=101 and then subtract by putting n=91 we get total combination = 176851 - 129766 = 47085 ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 2nd method) x1 + x2 + x3 + x4 =100 total combination = 100+4-1C100 = 103C100 = 176851 now to get the desired result subtract the case that value of x1 is atleast 11 11 + x1 + x2 + x3 + x4 = 100 x1 + x2 + x3 + x4 = 89 total combination = 89+4-1C89 = 92C89 = 125580 So ans = 176851 - 125580 = 51271 So which one is correct and why ? 0 votes 0 votes papesh commented Nov 5, 2016 reply Follow Share I have used second one. 47085 when u take atleast 10 instead of 11 that was my silly mistake when I have solved it first time.... 1 votes 1 votes Please log in or register to add a comment.