In how many ways can N students be seated in a row with N seats?
(1) |N − 6| = 3
(2) N^2 = 7N + 18
The OA is the option B.
I didn't understand this DS question. Can any expert clarify this for me? Thanks in advanced.
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In how many ways can N students be seated in a row with
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GMATGuruNY
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The number of ways to arrange N students = N!.M7MBA wrote:In how many ways can N students be seated in a row with N seats?
(1) |N − 6| = 3
(2) N^2 = 7N + 18
Thus, to answer the question stem, we need to know the value of N.
Question stem, rephrased:
What is the value of N?
Statement 1:
If N-6 = 3, then N=9.
If N-6 = -3, then N=3.
Since N can be different values, INSUFFICIENT.
Statement 2:
N² - 7N - 18 = 0
(N-9)(N+2) = 0
N=9 or N=-2.
Since the number of students must be POSITIVE, only N=9 is viable.
SUFFICIENT.
The correct answer is B.
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Vincen
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Hello.
Since there are N students and N seats in a row, the number of way to organize them is N!
Now, the question is, what is N?
Using statement (1) we get that N=3 or N=9. Since 3! and 9! are different, this statement is not sufficient.
Using the statement (2) we get: $$N^2=7N+18\ \Rightarrow\ \ N^2-7N-18=0\ \Rightarrow\ \ \ \left(N-9\right)\left(N+2\right)=0$$ This give us two values of N, but one of hem is negative (N=-2) which is not possible.
Hence, we get only one solution N=9. Therefore, there are 9! ways to seat the 9 students. SUFFICIENT.
Thus, the answer is B.
Since there are N students and N seats in a row, the number of way to organize them is N!
Now, the question is, what is N?
Using statement (1) we get that N=3 or N=9. Since 3! and 9! are different, this statement is not sufficient.
Using the statement (2) we get: $$N^2=7N+18\ \Rightarrow\ \ N^2-7N-18=0\ \Rightarrow\ \ \ \left(N-9\right)\left(N+2\right)=0$$ This give us two values of N, but one of hem is negative (N=-2) which is not possible.
Hence, we get only one solution N=9. Therefore, there are 9! ways to seat the 9 students. SUFFICIENT.
Thus, the answer is B.
