In an examination the maximum marks for each of the three papers are 50 each. Max marks for the fourth paper are 100. Find the number of ways in which the candidate cans score 60 percent marks in aggregate
I have solved this as 153c3-52c3-3*102c3=48076...but OA is given as 110551..
i considered cases...all three non -ve...12C2 ....one negative 11C2....two negative 10C2...and all three negative 9C2....adding all...202
I think it shud be 402..... ur approach is spot on but one no. can be negative in 3 ways and similarly 3 pairs of nos can be negative. therefore, 12C2+3*11C2+3*10C2+9C2= 402
if N=3^a*3^b, a and b are any two integers whose range is from 1 to 6 what is the probability that N is a multiple of 729. a. 5/18 b. 5/11 c. 13/11 d. 13/18 e. none
@swap747 a+b can take values from 2 to 12, but N will be a multiple only when a+b is 6,7,8,9,10,11,12. a+b=2 --1 way, a+b=3 --2 ways in this way total will be 55 (1+2+3+...11)
@nik25 dude....its not 55...u r correct till a+b =7 but for a+b= 8 its not 7, rather its 5 and it decreases from then i.e for a+b= 12...its a=6 and b=6 so 1way. in any case maximum number of sets possible are 36. so even if u get 55, some have to be redundant.