An auditor wants to conduct audits in 4 schools, of 5 different cities

Question

An auditor wants to conduct audits in 4 schools, of 5 different cities, in such a way that no two schools belong to same city. If in each city, there are 2 schools then in how many ways can the auditor select the schools for audit?

(A) 5
(B) 20
(C) 40
(D) 80
(E) 160

(A) 5
(B) 20
(C) 40
(D) 80
(E) 160

ScottTargetTestPrep · Accepted Answer

Solution:
 
There are 2C1 = 2 choices of schools for each of the 4 cities he will visit and there are 5C4 = 5 ways to choose the 4 cities. Therefore, there are a total of 2 x 2 x 2 x 2 x 5 = 80 ways to visit the schools.

Answer: D

sumitkrocks · Answer

Select 4 cities out of 5 and select 1 school in each of the 4 cities

5C4*(2C1)^4=5*16=80

Option D

luisdicampo · Answer

Deconstructing the Question

There are  5 cities , each with  2 schools . The auditor must select 4 schools such that no two schools come from the same city.

So we must select 4 different cities and then choose 1 school from each.

Step-by-step

First, choose 4 cities out of 5:

\binom{5}{4} = 5

Next, for each selected city, choose 1 school out of 2:

2 	imes 2 	imes 2 	imes 2 = 2^4 = 16

Multiply:

5 	imes 16 = 80

Answer D

luisdicampo · Answer

Deconstructing the Question

There are  5 cities , each with  2 schools . The auditor must select 4 schools such that no two schools come from the same city.

So we must select 4 different cities and then choose 1 school from each.

Step-by-step

First, choose 4 cities out of 5:

\binom{5}{4} = 5

Next, for each selected city, choose 1 school out of 2:

2 	imes 2 	imes 2 	imes 2 = 2^4 = 16

Multiply:

5 	imes 16 = 80

Answer D

Axwe7 · Answer

Why can't it be 10 x 8 x 6 x 4?

_ _ _ _

In the first audit, he has 10 options, and each subsequent option reduces by 2?

I agree with the other reasoning provided, but I just wanted to find the discrepancy between my thinking and the OE.

Bunuel · Answer

Your method counts the same selection many times because it treats different orders as different cases.

So, 10 * 8 * 6 * 4 must be divided by 4!.

That gives (10 * 8 * 6 * 4)/4! = 80.

So the issue is overcounting by order.

An auditor wants to conduct audits in 4 schools, of 5 different cities

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

An auditor wants to conduct audits in 4 schools, of 5 different cities

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards