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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (03:05) correct
59%
(02:26)
wrong
based on 70
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Ten students are waiting to board a school bus. "Student 1" wants to board before "Student 2" boards and "Student 2" wants to board after "Student 3" boards. The remaining students have no boarding preferences. What is the total number of ways in which all the students can board the bus?
(A) 120
(B) 720
(C) 1209600
(D) 1814400
(E) 3026640
Are You Up For the Challenge: 700 Level Questions
(A) 120
(B) 720
(C) 1209600
(D) 1814400
(E) 3026640
Are You Up For the Challenge: 700 Level Questions
There are 6 ways the 3 students could enter the bus with no constraints.
Given the constraints, there are 2 ways for them to enter:
132 and 312
So, 1/3 of the orderings are possible.
10!/3 = 1209600
This assumes that order matters, that the students are unique.
Students "preference" doesn't clarify this ambiguity.
Answers should be recast as factorials. No way GMAT actually would require all the multiplying.
Posted from my mobile device
Given the constraints, there are 2 ways for them to enter:
132 and 312
So, 1/3 of the orderings are possible.
10!/3 = 1209600
This assumes that order matters, that the students are unique.
Students "preference" doesn't clarify this ambiguity.
Answers should be recast as factorials. No way GMAT actually would require all the multiplying.
Posted from my mobile device
01 Jan 2022, 22:05
Kudos
Bookmarks
Since the 3 students want to board in order, we need to arrange the other 7 around them.
__ I __ III __ II __
a + b + c + d = 7
Here a, b, c and d can be any numbers from 0-7.
Therefore the total ways in which the students can board will be 10C3 - 120
Sine the 3 students can also board as __III__I__II__, the students can assign themselves in 120 ways in this way as well.
Also, the 7 students can arrange themselves in 7! ways.
So the total will be = (120+120)*7! = 1209600
__ I __ III __ II __
a + b + c + d = 7
Here a, b, c and d can be any numbers from 0-7.
Therefore the total ways in which the students can board will be 10C3 - 120
Sine the 3 students can also board as __III__I__II__, the students can assign themselves in 120 ways in this way as well.
Also, the 7 students can arrange themselves in 7! ways.
So the total will be = (120+120)*7! = 1209600
