Concept:Rank of a word is determined by counting all words that appear before it in alphabetical order, using permutations with repeated letters.
Formula:Number of permutations of
n items with
p identical items:
p!n!​.
Solution:Letters of "UDAYPUR" sorted: A, D, P, R, U, U, Y.
Total letters: 7, with two U's.
Words starting with A, D, P, or R: Fix one letter, arrange remaining 6 (two U's):
2!6!​=360 each.
Total:
4×360=1440.
Words starting with U: Fix U. Next letter in word is D, so fix UD. Then third letter can be A, P, R, U, Y. Count those with third letter before Y:
• Fix UDAP: remaining 3 letters (R, U, Y):
3!=6 words.
• Fix UDAR: remaining 3 letters (P, U, Y):
3!=6 words.
• Fix UDAU: remaining 3 letters (P, R, Y):
3!=6 words.
Total from UDAP, UDAR, UDAU:
6+6+6=18.
Now fix UDAY. Remaining letters: P, R, U. They give
3!=6 words. Among them, first two are UDAYPRU and UDAYPUR. So only 1 word precedes the target in this subgroup.
Words starting with UA: fix UA, remaining 5 letters (D, P, R, U, Y):
5!=120 words.
Total words before UDAYPUR:
1440+120+18+1=1579.
Rank =
1579+1=1580.
Answer:Rank = 1580, which corresponds to option C.