1 votes 1 votes suppose 5 people are to be randomly selected from a group of 20 individuals consisting of 10 married couples, and we want to determine P(N), the probability that the 5 chosen are all unrelated. (That is, no two are married to each other.)? himgta asked Nov 2, 2018 himgta 914 views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply himgta commented Nov 2, 2018 reply Follow Share my approach is first we select 1 person from 20,then for next we have only 18 choices(removing the one that is selected and his husband/wife) & same for remaining one! (20*18*16*14*12)/20C5 1 votes 1 votes Utkarsh Joshi commented Nov 2, 2018 reply Follow Share himgta YES! I guess it is the correct answer,Isn't it? 0 votes 0 votes Mk Utkarsh commented Nov 2, 2018 reply Follow Share himgta your approach is correct 0 votes 0 votes himgta commented Nov 2, 2018 reply Follow Share @Mk Utkarsh @Utkarsh Joshi Answer given as (10C5*32)/20C5 0 votes 0 votes Utkarsh Joshi commented Nov 2, 2018 reply Follow Share himgta YES answer given by them is correct. With the approach, u suggested we are even considering their order, but in this case we have to just select people not arrange. Divide your answer by 5! that's what they are saying. 0 votes 0 votes srestha commented Nov 2, 2018 reply Follow Share given ans is correct because, we can d it like this Here 10 men and 10 women 1) Select 5 men from 10 men 2) Select 4 men from 10 men and 1 woman from 6 women 3) Select 3 men from 10 men and 2 woman from 7 women and .......... Now that will be $\binom{10}{5}+\binom{10}{4}\times \binom{6}{1}+\binom{10}{3}\times \binom{7}{2}+\binom{10}{2}\times \binom{8}{3}+\binom{10}{1}\times \binom{9}{4}+\binom{10}{5}=32\times \binom{10}{5}$ 0 votes 0 votes himgta commented Nov 2, 2018 reply Follow Share @srestha mam @Utkarsh Joshi what is wrong with my approach? 0 votes 0 votes Utkarsh Joshi commented Nov 2, 2018 reply Follow Share As i said you are counting permutations which is not required. 0 votes 0 votes Utkarsh Joshi commented Nov 2, 2018 reply Follow Share According to your approach, ABCDE and BACDE are different but we just want the number of selections and not permutations. 0 votes 0 votes himgta commented Nov 3, 2018 reply Follow Share Thanks @Utkarsh Joshi brother! 0 votes 0 votes Please log in or register to add a comment.