Answer
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Hint: 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 1$
After multiplication we get $24$
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
$\left| \!{\underline {\,
3 \,}} \right. \times 2$(here we multiply by two because a and b can arrange in only 2 way)
$3 \times 2 \times 2$ (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
$\left| \!{\underline {\,
3 \,}} \right. \times 2$(here we multiply by two because c and d can arrange in only 2 way)
$3 \times 2 \times 2$ (opening of factorial)
So the number of arrangements is 12.
Now we find the number of ways where ab and cd both come together
$\left| \!{\underline {\,
2 \,}} \right. \times 2 \times 2$
$8$
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 = $24 - 12 - 12 + 8$
So answer is $8$
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.
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 1$
After multiplication we get $24$
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
$\left| \!{\underline {\,
3 \,}} \right. \times 2$(here we multiply by two because a and b can arrange in only 2 way)
$3 \times 2 \times 2$ (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
$\left| \!{\underline {\,
3 \,}} \right. \times 2$(here we multiply by two because c and d can arrange in only 2 way)
$3 \times 2 \times 2$ (opening of factorial)
So the number of arrangements is 12.
Now we find the number of ways where ab and cd both come together
$\left| \!{\underline {\,
2 \,}} \right. \times 2 \times 2$
$8$
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 = $24 - 12 - 12 + 8$
So answer is $8$
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.
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