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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

Explanation

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 1Aftermultiplicationweget After multiplication we get24So24istotalpossiblenumberofarrangementoflatterabcdNowtakingaandbas1whileNowwehavetotal3latterandarrangementofthesethreelatteris So 24 is total possible number of arrangement of latter ‘abcd’ Now taking a and b as 1 while Now we have total 3 latter and arrangement of these three latter is \left| \!{\underline {,
3 ,}} \right. \times 2(herewemultiplybytwobecauseaandbcanarrangeinonly2way)(here we multiply by two because a and b can arrange in only 2 way) 3 \times 2 \times 2(openingoffactorial)Sothenumberofarrangementsis12.Nowtakingcanddas1whileNowwehavetotal3latterandarrangementofthesethreelatteris(opening of factorial) So the number of arrangements is 12. Now taking c and d as 1 while Now we have total 3 latter and arrangement of these three latter is \left| \!{\underline {,
3 ,}} \right. \times 2(herewemultiplybytwobecausecanddcanarrangeinonly2way)(here we multiply by two because c and d can arrange in only 2 way) 3 \times 2 \times 2(openingoffactorial)Sothenumberofarrangementsis12.Nowwefindthenumberofwayswhereabandcdbothcometogether(opening of factorial) So the number of arrangements is 12. Now we find the number of ways where ab and cd both come together \left| \!{\underline {,
2 ,}} \right. \times 2 \times 2 8SonowwehavetofindnumberofwayswhereneitherabnorcdcometogetherisfindbygivenformulaTotalnumberofarrangementsnumberofarrangementofabnumberofarrangementofcd+numberofarrangementswhereabandcdtogetherSoneitherabnorcdcometogetheris= So now we have to find number of ways where neither ab nor cd come together is find by given formula Total number of arrangements – number of arrangement of ab - number of arrangement of cd + number of arrangements where ab and cd together So neither ab nor cd come together is =24 - 12 - 12 + 8Soansweris So answer is8$
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.