13as the ans 13*9C2[7]
There are 15 tickets numbered from 1 to 15 . Find the number ways in which 3 units having 5,3 and 2 tickets with consecutive numbers can be selected distributed amongst 3 children.
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8 Answers
13actually its p n c ... but styl help me out na.....
no abhirup ur ans not correct....
62first fix a, b, c
the first ticket number for each
b=a+5+x
c=b+3+y
c+1<=15
a>=1
hence
x,y>=0
c+1+3+5+x+y<=15
hence
a+x+y<=6
t=a+1
t+x+y<=5
divide 5 into three parts, 6C2
no of solutions is 7C2 x 3!
check for any counting mistakes :)
Corrected after MAK's clarification :)
62a, b, c are the starting positions of each of the series of tickets...
1bhaiya... d inequality should be
c+1=<15
then following d same procedure 4m that step... the final answer is...
7C2.3! = 126
i got d same answer by writing all possible sets (by some odd logics)... hence proved... [6] [3]