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14 Jan 2021, 03:45
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
87% (00:46) correct
13%
(01:21)
wrong
based on 202
sessions
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Time
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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!
A. 8/2
B. 6!
C. 6! * 2!
D. 8! - 2*7!
E. 8! - 2!
14 Jan 2021, 07:59
Kudos
Bookmarks
Bunuel
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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General Discussion
03 Mar 2021, 10:18
Kudos
Bookmarks
Bunuel
Out of eight alphabets 2 at the extreme ends are fixed.
Inner six are all different 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.
Answer B.
