Classrooms A has capacity of 25 seats. Ax denotes the number of possib

Question

Classrooms A has capacity of 25 seats.  A_x  denotes the number of possible seating arrangements of room ‘A’, when 25 of ‘x’ students are to be seated in this room in a row. If  A_n – A_{n–1} = 25!(49C25)  then what is the value of ‘n’?

(A) 51
(B) 50 
(C) 49 
(D) 48
(E) 47

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adityashinde88 · Answer

Hey  Bunuel  Sir, can you share the solution as well..

DrAnkita91 · Answer

please sir, can i get the solution of this? I am not understanding the question clearly

Kinshook · Answer

Given: Classrooms A has capacity of 25 seats.  A_x  denotes the number of possible seating arrangements of room ‘A’, when 25 of ‘x’ students are to be seated in this room in a row.
Asked: If  A_n – A_{n–1} = 25!(49C25)  then what is the value of ‘n’?

A_x = ^xC_{25}*25! 
 A_n - A_{n-1} = 25! (^nC_{25} - ^{n-1}C_{25}) = 25!^{49}C_{25} 
 ^nC_{25} - ^{n-1}C_{25} = \frac{n!}{25!(n-25)!} - \frac{(n-1)!}{25!(n-26)!} =\frac{ n!-(n-1)!(n-25)}{25!(n-25)! }= \frac{(n-1)! (n-(n-25))}{25!(n-25)!} = \frac{(n-1)!}{24!(n-1-24)!} = ^{n-1}C_{24} = ^{49}C_{25}

If n=50; 
 ^{49}C_{24} = ^{49}C_{25}

IMO B

chandy123 · Answer

Given:

Classroom A has a capacity of 25 seats. $ A_x $ denotes the number of possible seating arrangements when **25 of $ x $** students are seated in a row in room A. The problem gives the equation:

$ A_n - A_{n-1} = 25! \cdot \binom{49}{25} $

We need to find the value of $ n $.

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Step 1: Calculate $ A_n $ and $ A_{n-1} $

The number of seating arrangements when **25 students** are chosen from $ n $ total students is given by the permutation formula:

$ A_n = P(n, 25) = \frac{n!}{(n - 25)!} $ 
$ A_{n-1} = P(n-1, 25) = \frac{(n-1)!}{(n - 26)!} $

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Step 2: Simplify $ A_n - A_{n-1} $

We can rewrite $ A_n $ and $ A_{n-1} $ in terms of products:

$ A_n = n \cdot (n-1) \cdot (n-2) \cdots (n-24) $ 
$ A_{n-1} = (n-1) \cdot (n-2) \cdots (n-25) $

Taking the difference:

$ A_n - A_{n-1} = \left  \cdot (n-1)(n-2)\cdots(n-25) = 25 \cdot (n-1)(n-2)\cdots(n-25) $

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Step 3: Compare with the given equation

The given equation is:

$ A_n - A_{n-1} = 25! \cdot \binom{49}{25} $

We can equate the factors of $ 25! $ on both sides:

$ n = 49 $

Classrooms A has capacity of 25 seats. Ax denotes the number of possib

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GMAT Problem Solving (PS) Questions

Classrooms A has capacity of 25 seats. Ax denotes the number of possib

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