Question
Question: The number of arrangements of the latter ‘abcd’ in which neither ab nor cd come together is : (A) ...
The number of arrangements of the latter ‘abcd’ in which neither ab nor cd come together is :
(A) 6
(B) 12
(C) 16
(D) 8
Solution
This is a particular problem of permutation. Whenever questions talk that no two pairs of words come together, consider those as a single unit and find a number of arrangements. Permutations is the arrangement of the items whereas combination is the selection of the items among many items.
Complete step-by-step answer:
First step: we find the total possible arrangement of the latter ‘abcd’.
So we have a total four alphabet and the arrangement of all four alphabet is factorial of 4.
That is $$$$$\left| \!{\underline {,
4 ,}} \right. = 4 \times 3 \times 2 \times 1Aftermultiplicationweget24So24istotalpossiblenumberofarrangementoflatter‘abcd’Nowtakingaandbas1whileNowwehavetotal3latterandarrangementofthesethreelatteris\left| \!{\underline {,
3 ,}} \right. \times 2(herewemultiplybytwobecauseaandbcanarrangeinonly2way)3 \times 2 \times 2(openingoffactorial)Sothenumberofarrangementsis12.Nowtakingcanddas1whileNowwehavetotal3latterandarrangementofthesethreelatteris\left| \!{\underline {,
3 ,}} \right. \times 2(herewemultiplybytwobecausecanddcanarrangeinonly2way)3 \times 2 \times 2(openingoffactorial)Sothenumberofarrangementsis12.Nowwefindthenumberofwayswhereabandcdbothcometogether\left| \!{\underline {,
2 ,}} \right. \times 2 \times 28SonowwehavetofindnumberofwayswhereneitherabnorcdcometogetherisfindbygivenformulaTotalnumberofarrangements–numberofarrangementofab−numberofarrangementofcd+numberofarrangementswhereabandcdtogetherSoneitherabnorcdcometogetheris=24 - 12 - 12 + 8Soansweris8$
So option D is correct.
Note: Remember one thing whenever questions say about no two alphabet come together we have to make a pair and find all arrangements. The alphabets which come together must be selected as a single unit and the other items which do not come together can be taken as different units.