Question: 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)
- 8/2
- 6!
- 6! * 2!
- 8! - 2*7!
- 8! - 2!
Correct Answer: B
Solution and Explanation:
Approach Solution 1:
The problem statement asks to find the number of words that can be formed by rearranging the letters of the word PROBLEMS. It needs to be done in such a way that P and S occupy the first and last positions respectively.
Let us arrange the 8 letters and break them into stages.
Let us begin with the most restrictive stages.
Stage 1:
We will select a letter for the 1st position
Since the first letter must be P, we can complete stage 1 in 1 way
Stage 2:
We will 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:
Let us 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:
We will 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:
Let us select a letter for the 4th position
4 letters remaining.
So, we can complete stage 5 in 4 ways
Stage 6:
Now, we will select a letter for the 5th position
3 letters remaining.
So, we can complete stage 6 in 3 ways
Stage 7:
Selecting a letter for the 6th position
2 letters remaining.
So, we can complete stage 7 in 2 way
Stage 8:
Selecting 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)
Therefore, the number of words that can be formed by rearranging the letters of the word PROBLEMS = 6!
Approach Solution 2:
The problem statement asks to find the number of words that can be formed by rearranging the letters of the word PROBLEMS. It needs to be done in such a way that P and S occupy the first and last positions respectively.
Since 1st and last letters have only one way to form words, we cannot make new words. Out of eight alphabets, 2 at the extreme ends are fixed.
The inner six are all different and hence can be arranged in 6! ways.
=(ways for 2nd alphabet * 3rd alphabet * 4th alphabet ... 7th alphabet = 6*5*4*3*2*1)
Had P and S exchanged positions, the possible arrangements would have been 6! * 2.
Hence, the number of words that can be formed by rearranging the letters of the word PROBLEMS = 6!
Approach Solution 3:
The problem statement asks to find the number of words that can be formed by rearranging the letters of the word PROBLEMS. It needs to be done in such a way that P and S occupy the first and last positions respectively.
P can be positioned in the 1st place n 1 way.
S can be positioned in the last place in 1 way.
R, O, B, L, E, M can be arranged in between in 6! ways.
Therefore, the total number of words that can be formed by rearranging the letters of the word PROBLEMS = 1*1*6! = 6!
“How many words can be formed by re-arranging the letters of the word PROBLEMS”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This topic has been taken from the book “501 GMAT Questions”. The GMAT Problem Solving questions test the candidates' skills in analysing and solving quantitative problems. The candidates can promote their mathematical knowledge by practising more questions from the GMAT Quant practice papers to score better in the exam.
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