How many different words can be formed by rearranging the letters of the word error

Answer

Verified

Hint: Here, count the number of letters in the given word and also count the number of times a particular letter is repeated, and apply permutation formula to find the number of arrangements.

Complete step by step answer:
We are given the word “PATALIPUTRA” in which total number of letters are 11 (i.e. $P, A, T, A, L, I, P, U, T, R, A$).
Total number of vowels = $5 (i.e. 3 A’s, 1 I $ and $1 U)$
Total number of consonants = $6 (i.e. 2 P’s, 2 T’s, 1 L $ and $1 R)$
Also, given that words can be formed with the letters of the word PATALIPUTRA without changing the relative positions of vowels and consonants.
So, total number of words = $\dfrac{{5!}}{{3!}} \times \dfrac{{6!}}{{2!2!}}$
[Permutation formula: In arrangement of n letters in which letters a and b are repeated $x$ and $y$ times, then the number of possible arrangements is given as $\dfrac{{n!}}{{x!y!}}$.
Here, $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$5! = 5 \times 4 \times 3 \times 2 \times 1$
$3! = 3 \times 2 \times 1$
$2! = 2 \times 1$
Putting all values and simplifying, we get
Total number of words = $4 \times 5 \times 180 = 3600$

Therefore, the number of words that can be formed with the letters of the word $\text{PATALIPUTRA}$ without changing the relative positions of vowels and consonants is 3600. Hence, option (C) is correct.

Note:
In these types of question, first check whether question is asked about combination or
permutation. Permutation means arrangement of things, and combination means taking a particular number of items at a time (arrangement does not matter in combination). Then apply the proper formula as required. Observe that the given condition is with or without repetition condition and proceed for error free calculations.

Re: How many words can be formed by re-arranging the letters of the word [#permalink]

Bunuel wrote:

How many words can be formed by re-arranging the letters of the word PROBLEMS such that P and S occupy the first and last position respectively? (Note: The words thus formed need not be meaningful)

A. 8/2
B. 6!
C. 6! * 2!
D. 8! - 2*7!
E. 8! - 2!

Take the task of arranging the 8 letters and break it into stages.

Begin with the most restrictive stages.

Stage 1: Select a letter for the 1st position
Since the first letter must be P, we can complete stage 1 in 1 way

Stage 2: Select a letter for the last (8th) position
Since the last letter must be S, we can complete stage 2 in 1 way

Stage 3: Select a letter for the 2nd position
There are 6 remaining letters from which to choose
So, we can complete stage 3 in 6 ways

Stage 4: Select a letter for the 3rd position
There are 5 remaining letters from which to choose
So, we can complete stage 4 in 5 ways

Stage 5: Select a letter for the 4th position
4 letters remaining. So, we can complete stage 5 in 4 ways

Stage 6: Select a letter for the 5th position
3 letters remaining. So, we can complete stage 6 in 3 ways

Stage 7: Select a letter for the 6th position
2 letters remaining. So, we can complete stage 7 in 2 way

Stage 8: Select a letter for the 7th position
1 letter remaining. So, we can complete stage 8 in 1 way

By the Fundamental Counting Principle (FCP), we can complete all 8 stages (and thus seat all 6 children) in (1)(1)(6)(5)(4)(3)(2)(1) ways (= 6! ways)

Answer: B

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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