The problem asks for the number of 4-letter words that can be formed using the letters P, Q, R, P, Q, R, S, T, U, V, P. First, let's count the frequency of each letter available:
- P: 3 times
- Q: 2 times
- R: 2 times
- S: 1 time
- T: 1 time
- U: 1 time
- V: 1 time
There are 7 distinct letters: P, Q, R, S, T, U, V.
We need to form 4-letter words. We consider different cases based on the repetition of letters.
Case 1: All 4 letters are distinct
We need to choose 4 distinct letters from the 7 available distinct letters (P, Q, R, S, T, U, V) and arrange them.
Number of ways = Permutation of 7 items taken 4 at a time.
$ P(7, 4) = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840 $
Case 2: One pair of identical letters and 2 distinct letters
First, choose the letter that will form the pair. The letters available for pairing are P (3 times), Q (2 times), R (2 times).
- Subcase 2a: Pair is PP. Choose 2 distinct letters from the remaining 6 distinct letters (Q, R, S, T, U, V). Number of ways = $^6C_2 = \frac{6 \times 5}{2 \times 1} = 15$. The arrangement is of the type PPXY. Number of arrangements = $\frac{4!}{2!} = \frac{24}{2} = 12$. Total ways = $15 \times 12 = 180$.
- Subcase 2b: Pair is QQ. Choose 2 distinct letters from the remaining 6 distinct letters (P, R, S, T, U, V). Number of ways = $^6C_2 = 15$. The arrangement is of the type QQXY. Number of arrangements = $\frac{4!}{2!} = 12$. Total ways = $15 \times 12 = 180$.
- Subcase 2c: Pair is RR. Choose 2 distinct letters from the remaining 6 distinct letters (P, Q, S, T, U, V). Number of ways = $^6C_2 = 15$. The arrangement is of the type RRXY. Number of arrangements = $\frac{4!}{2!} = 12$. Total ways = $15 \times 12 = 180$.
Total ways for Case 2 = $180 + 180 + 180 = 540$.
Case 3: Two pairs of identical letters
Choose 2 letters from {P, Q, R} which can form pairs. Number of ways = $^3C_2 = 3$. The possible pairs are (PP, QQ), (PP, RR), (QQ, RR).
For each combination (e.g., PPQQ), the number of arrangements is $\frac{4!}{2!2!} = \frac{24}{4} = 6$.
Total ways for Case 3 = $3 \times 6 = 18$.
Case 4: Three identical letters and 1 distinct letter
The only letter available 3 times is P. So, the word must contain PPP.
Choose 1 distinct letter from the remaining 6 distinct letters (Q, R, S, T, U, V). Number of ways = $^6C_1 = 6$. The arrangement is of the type PPPX.
Number of arrangements = $\frac{4!}{3!} = \frac{24}{6} = 4$.
Total ways for Case 4 = $6 \times 4 = 24$.
Total Number of Words
Sum the results from all cases:
Total = Case 1 + Case 2 + Case 3 + Case 4
Total = $840 + 540 + 18 + 24 = 1422$.
Similar Questions
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
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- Let $z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$, where $i = \sqrt{-1}$. If $|z|^2 = 44200$, then $n$ is equal to _______
- Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is q, then $p + q$ is equal to ____________.
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If $ \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)$ $\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)$ $\dots$ $ \left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right)$ = $\frac{\alpha^{13}}{^{14}C_0 ^{14}C_1 \dots ^{14}C_{12}} $, then $30\alpha$ is equal to __________.
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Important Questions from Algebra
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$, where $i = \sqrt{-1}$. If $|z|^2 = 44200$, then $n$ is equal to _______
- Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is q, then $p + q$ is equal to ____________.
