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There are 5 doors to a lecture room. Two are red and the oth
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08 May 2019, 04:26
08 May 2019, 04:26
Expert Reply
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Question Stats:
65% (01:01) correct
34% (00:46) wrong based on 26 sessions
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There are 5 doors to a lecture room. Two are red and the others are green. In how many ways can a lecturer enter the room and leave the room from different colored doors?
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
(A) 1
(B) 3
(C) 6
(D) 9
(E) 12
Re: There are 5 doors to a lecture room. Two are red and the oth
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12 May 2019, 20:23
12 May 2019, 20:23
1
We can see that 2 red doors are identical and 3 green doors are identical. R=2 and G=3
Now the combination part comes as if we enter red door first - selecting one door to enter (red) from two possible doors and then selecting one door to exit( green) from three possible green doors - 2C1*3C1 will be case 1
If we enter from the Green door - selecting one door to enter (Green) from possible doors (Green) then selecting one door to exit (red) from two possible doors( Red) - 3C1*2C1 will be case 2
Total makes : 2C1*3C1+3C1*2C1 = 6+6= 12
Now the combination part comes as if we enter red door first - selecting one door to enter (red) from two possible doors and then selecting one door to exit( green) from three possible green doors - 2C1*3C1 will be case 1
If we enter from the Green door - selecting one door to enter (Green) from possible doors (Green) then selecting one door to exit (red) from two possible doors( Red) - 3C1*2C1 will be case 2
Total makes : 2C1*3C1+3C1*2C1 = 6+6= 12
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Location: United States
GMAT 1: 770 Q51 V44
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WE:Education (Education)
Re: There are 5 doors to a lecture room. Two are red and the oth
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13 May 2019, 07:26
13 May 2019, 07:26
3
Expert Reply
To avoid using Combination or Permutation notation, you can always just consider the following steps to manually calculate the scenario:
1) Draw a ____ for Each Selection to be Made and Label any Dictated Restrictions for the Selection.
In this Case there will be a selection of two doors:
Door 1 and Door 2
2) Determine Most Efficient Method to Properly and Specifically Make Selection based on Problem and Consider Whether Selecting this or That.
In this Case the First door Selection Impacts the Second Door Selection So We Need to Split into Two Scenarios:
Door 1 (Red) and Door 2 (Green)
OR
Door 1 (Green) and Door 2 (Red)
3) Then Insert the Number or Available Options to Select at Moment of Selection One Blank at a Time
In this Case for the first iteration:
2 options (Red) and 3 options (Green)
OR
3 options (Green) and 2 options (Red)
4) Finally, calculate. When Selecting This AND That - Multiply. So, in the first iteration there are 2 x 3 = 6 ways to enter a Red door and exit a Green door as well as 3 x 2 = 6 ways to enter a Green door and exit a Red door in the second iteration. Then, when it is possible to Select this OR That - Add. So, add the 6 ways to proceed Red then Green to the 6 ways to proceed Green then Red to determine that there are 12 total ways to enter one color door and exit the other.
1) Draw a ____ for Each Selection to be Made and Label any Dictated Restrictions for the Selection.
In this Case there will be a selection of two doors:
Door 1 and Door 2
2) Determine Most Efficient Method to Properly and Specifically Make Selection based on Problem and Consider Whether Selecting this or That.
In this Case the First door Selection Impacts the Second Door Selection So We Need to Split into Two Scenarios:
Door 1 (Red) and Door 2 (Green)
OR
Door 1 (Green) and Door 2 (Red)
3) Then Insert the Number or Available Options to Select at Moment of Selection One Blank at a Time
In this Case for the first iteration:
2 options (Red) and 3 options (Green)
OR
3 options (Green) and 2 options (Red)
4) Finally, calculate. When Selecting This AND That - Multiply. So, in the first iteration there are 2 x 3 = 6 ways to enter a Red door and exit a Green door as well as 3 x 2 = 6 ways to enter a Green door and exit a Red door in the second iteration. Then, when it is possible to Select this OR That - Add. So, add the 6 ways to proceed Red then Green to the 6 ways to proceed Green then Red to determine that there are 12 total ways to enter one color door and exit the other.
