6 votes 6 votes How many ways are there to Distribute 7 distinct objects to 3 Distinct boxes and No box should be Empty Any box can be Empty Combinatory combinatory counting + – [ Jiren ] asked Aug 22, 2022 • retagged Aug 22, 2022 by makhdoom ghaya [ Jiren ] 477 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Kabir5454 commented Aug 22, 2022 i moved by Kabir5454 Aug 23, 2022 reply Follow Share Part 1:- No box should be empty implies each box should have at least one object . This is same as number onto function $f:A->B$ where $|A|=7$ and $|B|=3$ Number of onto function = $3^{7}-3*2^{7}+3=1806$ Part 2:- Number of arrangement where any box can be empty= total number of arrangement $-$ number of arrangement with no box is empty Number of arrangement where any box can be empty= $3^{7}-1806=381$ 5 votes 5 votes [ Jiren ] commented Aug 22, 2022 reply Follow Share In part 2 Any of the box can be empty can be treated as No of functions from A → B right so 3^7 should be the answer its asked “ any box CAN be empty ” means there are no restrictions in the distribution right 6 votes 6 votes Kabir5454 commented Aug 22, 2022 reply Follow Share The wording “any box can be empty “ is ambiguous to me . As per me i have given solution based on at least one box is empty . Your reasoning is also seems correct to me . 0 votes 0 votes robinofautumn commented Oct 19, 2022 reply Follow Share this answer seems to be more convincing 0 votes 0 votes Riya_23 commented Dec 21, 2022 reply Follow Share any box can be empty vs some box must be empty...so part 2 is not ambiguous 0 votes 0 votes Please log in or register to add a comment.