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A coach arranged 6 players, each with different height in two parallel
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12 Sep 2021, 21:31
12 Sep 2021, 21:31
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Question Stats:
54% (01:46) correct
45% (01:46) wrong based on 44 sessions
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A coach arranged 6 players, each with different height in two parallel rows of three players each. Heights of the player within each row increase from left to right, and each player in the second row must be taller than the person standing in front of him or her.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Total number of possible arrangements |
6 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: A coach arranged 6 players, each with different height in two parallel
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13 Sep 2021, 03:21
13 Sep 2021, 03:21
C?
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Posted from my mobile device
A coach arranged 6 players, each with different height in two parallel
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13 Sep 2021, 10:24
13 Sep 2021, 10:24
2
Very cool question!
For this one, I would start by numbering the players 1-6 respectively.
1 2 3
4 5 6
How many times can we permute this matrix within the constraints?
Before we start, it's important to note that player 1 and player 6 can never move positions, otherwise, the left-to-right constraint will be violated.
So we begin by assessing how we can create the top row.
We have:
1 2 3
or
1 2 4
or
1 2 5
or
1 3 4
or
1 3 5
You'll see that the bottom row fills itself in each example.
The question now becomes, can we create any more top rows?
For example, can we create 1 4 5?
The answer is no, because the only numbers greater than 4 are 5 and 6, and 5 is already in the top row and remember, 6 cannot move from that bottom row spot.
Therefore, the maximum number of arrangements is 5, so the answer is B
For this one, I would start by numbering the players 1-6 respectively.
1 2 3
4 5 6
How many times can we permute this matrix within the constraints?
Before we start, it's important to note that player 1 and player 6 can never move positions, otherwise, the left-to-right constraint will be violated.
So we begin by assessing how we can create the top row.
We have:
1 2 3
or
1 2 4
or
1 2 5
or
1 3 4
or
1 3 5
You'll see that the bottom row fills itself in each example.
The question now becomes, can we create any more top rows?
For example, can we create 1 4 5?
The answer is no, because the only numbers greater than 4 are 5 and 6, and 5 is already in the top row and remember, 6 cannot move from that bottom row spot.
Therefore, the maximum number of arrangements is 5, so the answer is B
