A Team of 8 Students Goes On An Excursion, In Two Cars, Of Which One GMAT Problem Solving

‘A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4’ is the topic from the GMAT Quantitative problem set. GMAT quantitative reasoning section tests the candidate's ability to solve mathematical and quantitative problems, interpret graph data, and mathematical reasoning. The question in this section comes with five options. Candidates need to choose the one which is correct. GMAT Quant syllabus has mainly the two categories-

  • Problem Solving: This question type in GMAT Quantitative analyses candidates logical and analytical reasoning skills. In this section, candidates indicate the best five answer choices.
  • Data Sufficiency: This question type in GMAT Quantitative analyses candidates’ quantitative problems and identifies relevance with the data given.

Topic: A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. If internal arrangement inside the car does not matter then the number of ways in which they can travel, is

  1. 91
  2. 122
  3. 126
  4. 3920
  5. 4120

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Answer: C

Model Answer:

There is only one approach to solve the problem.

Explanation:
Given:

  • 8 students goes on an excursion, in two cars
  • one can seat 5 and the other only 4.

Condition:

  • The internal arrangement inside the car does not matter

Find out:

  • The number of ways in which they can travel

Let’s first assume that the students are indistinguishable.
Then the problem becomes: How many ways can 2 positive integers (where the first addend can’t be more than 4 and the second can’t be more than 5) add up to 8?
There are two ways we can attain this:

  1. 3 + 5 = 8
  2. 4 + 4 = 8

Now, obviously the students (and in general, people) are distinguishable, and they can exchange seats. Hence, the internal arrangement will change. Considering this, we have to take each of the two cases above:
Considering Case 1: 3 + 5 = 8
That is, the first car gets 3 students and the second 5 students. Since the students are distinguishable, the number of choices the first car has is 8C3 and the number of choices the second car has is 5C5 (after 3 students have been chosen to sit in the first car).
Therefore, the number of ways the 8 students can be distributed into 2 cars in this case is
8C3 x 5C5
= (8 x 7 x 6)/(3 x 2) x 1
= 8 x 7
= 56.
Considering Case 2: 4 + 4 = 8
That is, the first car gets 4 students and the second 4 students. Since the students are distinguishable, the number of choices the first car has is 8C4 and the number of choices the second car has is 4C4 (after 4 students have been chosen to sit in the first car).
Therefore, the number of ways the 8 students can be distributed into 2 cars in this case is
8C4 x 4C4
= (8 x 7 x 6 x 5)/(4 x 3 x 2) x 1
= 2 x 7 x 5
= 70.
Thus, the total number of ways in both cases is 56 + 70 = 126.
Hence, C is the correct answer.

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