2 votes 2 votes The number of ways in which n distinct objects can be put into two identical boxes so that no box remains empty, is a) 2^n - 1 b) 2^n - 2 c) 2^(n-1) - 1 d) None of these Please explain your answer. Combinatory combinatory discrete-mathematics + – Jatin18 asked Jun 11, 2017 • retagged Jun 27, 2017 by Arjun Jatin18 1.4k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply sid1221 commented Jun 11, 2017 reply Follow Share is it c? i just try with example 0 votes 0 votes Jatin18 commented Jun 11, 2017 reply Follow Share yes the ans is c. 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes S(n,2) = 2^(n-1) - 1 http://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php Follow this link, to know it in detail. Ahwan answered Jun 11, 2017 • edited Jun 16, 2017 by Ahwan Ahwan comment Share Follow See all 6 Comments See all 6 6 Comments reply Jatin18 commented Jun 11, 2017 reply Follow Share Got it.2^n - 2 ways if the boxes are not identical. But why did you subtract one for identical boxes case.Shouldn't we divide by two for identical boxes case to counter case which now will be similar due to identical boxes. 0 votes 0 votes Ahwan commented Jun 11, 2017 reply Follow Share I have edited the answer. See. 0 votes 0 votes Pinaki Dash commented Jun 12, 2017 reply Follow Share Doubt in a similar question. How many ways are there to distribute five distinguishable objects into three indistinguishable boxes? Can we solve it like following? First, we calculate the no of ways of distributing 5 distinct objects in 3 distinct boxes which are = $3^{5}$ and then we divide it with 3! as we can label the boxes in 3! ways. Final answer=$3^{5}$ /3! I know that using the formula of Stirling numbers of the second kind, it can be solved as S(5,1)+S(5,2)+S(5,3). But can the method stated above be used? 1 votes 1 votes Jatin18 commented Jun 12, 2017 reply Follow Share In S(k,n) what is S.How do we calculate such expression? 0 votes 0 votes Pinaki Dash commented Jun 12, 2017 reply Follow Share In S(k,n), S is the Stirling number of the second kind Stirling numbers of the second kind obey the recurrence relation S(k,n)= 1, if k>0 and n=1 1, if k=n n*S(k-1,n)+S(k-1,n-1), 0<=n<=k 1 votes 1 votes tirth_patel commented Sep 28, 2021 reply Follow Share https://www.cse.iitd.ac.in/~mittal/stirling.html 0 votes 0 votes Please log in or register to add a comment.