Questions & Answers

Question

Answers

(A) 6

(B) 12

(C) 16

(D) 8

Answer
Verified

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.