Question

If the permutations 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

The correct option is D $93$

Number of words beginning with $A=$ number of arrangements of $B,C,D,E=4!$
Similarly, number of words beginning with $B,C$ are $4!$ each.
Number of arrangement of words staring with $DA,DB,DC=3!$
Similarly, number of words beginning with $DB,DC$ are $3!$ each.
The next words are $DEABC,DEACB,DEBAC,...$
$\therefore$ Rank of $DEBAC=3(4!)+3(3!)+3=93$.