Question

If the permutation of $a,b,c,d,e$ taken all together be written down in alphabetical Order as in dictionary and numbered, then the rank of the permutation $debac$ is:
(a) $90$
(b) $91$
(c) $92$
(d) $93$

Answer 1

Hint: To solve the above question, we have to use the concept of permutation. We have to find the rank of ‘debac’. We can see that number of words beginning with $a=$ number of arrangements of $b,c,d,e=4!$, And similarly, we can see that number of words beginning with $b,c$ are $4!$ each. In this way, we have to
Solve this problem.

Complete step-by-step solution:
We know that in a dictionary, the words are listed and ranked in alphabetical order. So, in the given we problem have to find the rank of the word $'debac'$.
Now for finding the number of words we have starting with $a$, we have to find the number of arrangements of the remaining 4 letters.
Now, the number of such arrangements is =$4!$
Now we are finding the number of words starting with $b$, we have to find the number of arrangements of the remaining $4$ letters.
Now, the number of such arrangements is =$4!$
Now we are finding the number of words starting with $c$, we have to find the number of arrangements of the remaining $4$ letters.
Now, the number of such arrangements is =$4!$
Now we are finding the number of words starting with $d$ and we are fixing the next letter as $a$, we have to find the number of arrangements of the remaining $3$ letters.
Now, the number of such arrangements is =$3!$
Now we are finding the number of words starting with $d$ and we are fixing the next letter as $b$, we have to find the number of arrangements of the remaining $3$ letters.
Now, the number of such arrangements is =$3!$
Now we are finding the number of words starting with $d$ and we are fixing the next letter as $c$, we have to find the number of arrangements of the remaining $3$ letters.
Now, the number of such arrangements is =$3!$
Now we are finding the number of words starting with $d$, fixing the next letter as e:
Where the first word -$deabc$
And the second word-$deacb$
And the third word-$debac$
So, we can see that number of words which we reach the word $debac=4!+4!+4!+3!+3!+3!+1+1+1=93$
Hence, the correct option is (d) $93$.

Note: Here students must take care of the concept of permutation. Sometimes, the student did a mistake between permutation and combination because they are different. We have to know the main difference between Permutation and combination is ordering. So, students have to take care of it.