Permutation of Repeated Objects

The word given is PQRPQRSTUVP. Here, we have P 3 times, Q 2 times, R 3 times, and S,T,U,V one time each
Total distinct letters: 7

We form 4-letter words using these.

Case 1: All letters are distinct

Here, 4 distinct letters can be chosen from 7 letters in  $$ {}^7C_4 = 7 \cdot 6 \cdot 5 \cdot 4 = 840 \text { ways} $$

Case 2: One pair + two distinct

To choose a pair, we need to pick one from the letters that occur 2 times or more in the word. Of the letters P,Q and R occure more than once, so the letter that will form the pair can be picked in 3 ways. For the remaining 2 distinct letters, they can be picked from the remaining 6 letters in  $$  \binom{6}{2} = 15 $$ ways. These can be arranged amongst themselves in  $$  \frac{4!}{2!} = 12$$ ways

Total number of ways words can be formed is= $$ 3 \cdot 15 \cdot 12 = 540 \text { ways} $$

Case 3: Two pairs

Here, we need to choose 2 letters (each must occur ≥2 times): from (P,Q,R), which can be done in 3 ways. The letters can be arranged amongst themselves in $$ \dfrac{4!}{2!2!} = 6 $$ ways

Total number of ways words can be formed is $$ 3 \cdot 6 = 18 \text{ ways} $$

Case 4: Three same + one different

Only P occurs three times, so here it can be picked in 1 way, and the 1 different letter from the remaining 6 can be picked in 6 ways. The letters can be arranged amongst themselves in  $$ \dfrac{4!}{3!} = 4  $$

Total number of ways words can be formed is $$ 1 \cdot 6 \cdot 4 = 24 \text{ ways}$$

Hence, the total number of ways words can be formed considering all the cases is  $$ 840 + 540 + 18 + 24 =1422 \text { ways} $$

When the letters of the word "UDAYPUR" are arranged in alphabetical order, we get A, D, P, R, U, U, Y.

We need to find the rank of the word "UDAYPUR" out of all the words formed by arranging the above letters.

Number of words formed with first letter as A$$=\dfrac{6!}{2!}$$ (because all the 6 letters can be arranged in 6! ways and there are 2 U's, so dividing by 2!)

Number of words formed with first letter as D$$=\dfrac{6!}{2!}$$ (because all the 6 letters can be arranged in 6! ways and there are 2 U's, so dividing by 2!)

Number of words formed with first letter as P$$=\dfrac{6!}{2!}$$ (because all the 6 letters can be arranged in 6! ways and there are 2 U's, so dividing by 2!)

Number of words formed with first letter as P$$=\dfrac{6!}{2!}$$ (because all the 6 letters can be arranged in 6! ways and there are 2 U's, so dividing by 2!)

Number of words formed with first two letters as UA $$=5!$$ (because all the 5 letters left can be arranged in 5! ways)

Number of words formed with first four letters as UDAP $$=3!$$ (because all the 3 letters left can be arranged in 3! ways)

Number of words formed with first four letters as UDAR $$=3!$$ (because all the 3 letters left can be arranged in 3! ways)

Number of words formed with first four letters as UDAU $$=3!$$ (because all the 3 letters left can be arranged in 3! ways)

Number of words formed with the first six letters as UDAYPR $$=1$$ (which is UDAYPRU)

All the above words precede the word "UDAYPUR" when arranged in a dictionary, and to get the rank, we need to add all the above values and 1 to it, since it is the next word.

The next word formed is "UDAYPUR", and the rank of it can be calculated as,

Rank $$=\dfrac{6!}{2!}+\dfrac{6!}{2!}+\dfrac{6!}{2!}+\dfrac{6!}{2!}+5!+3!+3!+3!+1+1=360\times\ 4+120+6\times\ 3+2=1580$$

So, the rank of the word is 1580.

Hence, the correct answer is option A.

Calculating x:

In this case, we have to form a 9-digit number where only one digit is repeated, and it is repeated exactly twice.

Selecting the repeated digit: $$^9C_1=9$$ ways

Now, one digit is used twice, we need 9 - 2 = 7 more distinct digits from the remaining 8 digits in S. This can be selected in $$^8C_7=8$$ ways.

Now, all these 9 digits can be arranged in $$\dfrac{9!}{2!}$$ ways. 

Total possible arrangements: $$9\times8\times\dfrac{9!}{2!}=36\times9!$$ which is equal to x. 

Calculating x:

In this case, we have to form a 9-digit number where two digits are repeated, and each is repeated exactly twice.

Selecting the repeated digits: $$^9C_2=36$$ ways

Now, two digits is used twice, we need 9 - 4 = 5 more distinct digits from the remaining 7 digits in S. This can be selected in $$^7C_5=21$$ ways.

Now, all these 9 digits can be arranged in $$\dfrac{9!}{2!2!}$$ ways.

Total possible arrangements: $$36\times21\times\dfrac{9!}{2!2!}=9\times21\times9!=189\times9!$$ which is equal to y.

To find the relationship, we can divide y by x. 

$$\dfrac{y}{x}=\dfrac{189\times9!}{36\times9!}$$

$$\dfrac{y}{x}=\dfrac{21}{4}$$

Or, we can say,  4y = 21x