Question 357242


Total no. of possible subsets of 5 elements (including empty set)

 = 5C0 + 5C1 + 5C2 + 5C3 +5C4 + 5C5 =  1 + 5 +10 + 10 + 5 + 1 =  32


a). no. of subset to ensure at least 4 identical subsets =  32*3 + 1


 = 97




b). 161 = 32 * 5  + 1


so, at least 6 identical subsets are printed.




Sometimes there may be typing mistake in solution of a problem, so please ignore it. Understand the concept and try to solve the problem yourself. If there is problem related to concept, contact at 
sudhanshu.cochin@yahoo.com or sudhanshu.cochin@gmail.com 
Best of luck.......