Definition of permutation, Number of permutations with or without repetition, Conditional permutations Questions in English

(C) The letters of the word "$UDAYPUR$" are $A, D, P, R, U, U, Y$. Total letters = $7$. The letter $U$ repeats $2$ times.
Alphabetical order: $A, D, P, R, U, Y$.
To find the rank, we count the number of words that come before "$UDAYPUR$":
$1$. Words starting with $A$: $\frac{6!}{2!} = 360$
$2$. Words starting with $D$: $\frac{6!}{2!} = 360$
$3$. Words starting with $P$: $\frac{6!}{2!} = 360$
$4$. Words starting with $R$: $\frac{6!}{2!} = 360$
$5$. Words starting with $UA$: $5! = 120$
$6$. Words starting with $UDAP$: $3! = 6$
$7$. Words starting with $UDAR$: $3! = 6$
$8$. Words starting with $UDAYP$: $1! = 1$
$9$. Words starting with $UDAYR$: $1! = 1$
$10$. Words starting with $UDAYU$: $1! = 1$
$11$. The next word is "$UDAYPUR$",which is $1$.
Total rank = $360 + 360 + 360 + 360 + 120 + 6 + 6 + 1 + 1 + 1 + 1 = 1576$.
Therefore, the rank of the word "$UDAYPUR$" is $1577$ (as the word itself is the $1577^{th}$ word).

(B) The word '$PQRPQRSTUVP$' contains $11$ letters: $P(3), Q(3), R(2), S(1), T(1), U(1), V(1)$.
Wait, let us re-count: $P, Q, R, P, Q, R, S, T, U, V, P$.
Letters are: $P: 3, Q: 2, R: 2, S: 1, T: 1, U: 1, V: 1$. Total distinct letters are $7$ $(P, Q, R, S, T, U, V)$.
Case $I$: $3$ alike, $1$ different: Choose $1$ letter from ${P}$ to be $3$ alike, and $1$ from remaining $6$ letters. Number of ways $= {}^{1}C_{1} \times {}^{6}C_{1} \times \frac{4!}{3!} = 1 \times 6 \times 4 = 24$.
Case $II$: $2$ alike, $2$ alike: Choose $2$ letters from ${P, Q, R}$ to be $2$ alike each. Number of ways $= {}^{3}C_{2} \times \frac{4!}{2!2!} = 3 \times 6 = 18$.
Case $III$: $2$ alike, $2$ different: Choose $1$ letter from ${P, Q, R}$ to be $2$ alike, and $2$ from remaining $6$ letters. Number of ways $= {}^{3}C_{1} \times {}^{6}C_{2} \times \frac{4!}{2!} = 3 \times 15 \times 12 = 540$.
Case $IV$: All $4$ different: Choose $4$ letters from $7$ distinct letters. Number of ways $= {}^{7}C_{4} \times 4! = 35 \times 24 = 840$.
Total words $= 24 + 18 + 540 + 840 = 1422$.

(D) Total number of people = $4 \text{ boys} + 3 \text{ girls} = 7 \text{ people}$.
Total ways to arrange $7$ people in a queue = $7! = 5040$.
To find the number of ways where all girls are not together, we use the complement method: $\text{Total ways} - \text{Ways where all girls are together}$.
Treating the $3$ girls as a single unit, we have $4 \text{ boys} + 1 \text{ unit} = 5 \text{ units}$.
These $5$ units can be arranged in $5!$ ways, and the $3$ girls within their unit can be arranged in $3!$ ways.
Ways where all $3$ girls are together = $5! \times 3! = 120 \times 6 = 720$.
Therefore, the number of ways where all girls are not together = $5040 - 720 = 4320$.

(D) The word $INCONSEQUENTIAL$ consists of $15$ letters.
First, identify the distinct vowels and consonants.
Vowels: {$I$, $O$, $E$, $U$, $A$} ($5$ distinct vowels).
Consonants: {$N$, $C$, $S$, $Q$, $T$, $L$} ($6$ distinct consonants).
We need to select $2$ vowels out of $5$ and $2$ consonants out of $6$.
The number of ways to choose these letters is $\binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150$.
Each selection contains $4$ distinct letters, which can be arranged in $4! = 24$ ways.
Total number of words = $150 \times 24 = 3600$.

(B) The string "$KCET$ $2025$" contains $8$ characters: $K, C, E, T, \text{space}, 2, 0, 2, 5$. However, assuming the space is ignored and we consider the $8$ characters: $K, C, E, T, 2, 0, 2, 5$.
Fix $K$ at the first position and $5$ at the last position.
The remaining $6$ characters are $C, E, T, 2, 0, 2$.
Among these $6$ characters, the digit $2$ repeats twice.
The number of ways to arrange these $6$ characters is $\frac{6!}{2!} = \frac{720}{2} = 360$ ways.