How many different ways can 2 students be seated in a row of

Question

How many different ways can 2 students be seated in a row of 4 desks, so that there is always at least one empty desk between the students?

(A) 2
(B) 3
(C) 4
(D) 6
(E) 12

honeyrai · Accepted Answer

Here order is important, therefore, we'll use Perm.

All the possible ways in which 2 stud can be seated on 4 seats = 4P2.
Suppose, two students are always seated together. Then we'll treat them as 1 stud & the seats left to be filled are 3. Therefore no. of ways in which 2 stud are always seated together = 3P2.

Now, the no. of ways in which 2 students have a gap in between = total no. of ways - no. of ways in which students are seated together = 4P2-3P2 = 12-6 = 6

LevFin7S · Answer

3 ways to seat the students:
with two empty seats between
1 empty w/ one student on the left most
1 empty....right most

two students can be interchanged

3x2=6

ksharma12 · Answer

But that yields the same answer as 4C2? How do you solve use combinations or perm?

That comb above does not take into account the empty seat.

gurpreetsingh · Answer

Take two cases.

case 1... 2 spaces in between that means C1 _ _ C2 or C2_ _ C1

case 2.... 1 space in between.
now consider c1_c2 as one element
Number of ways of arranging these in 4 seats =  2p2*2!  = 4

2p2 because we have considered c1_c2 as one element , 2! is rearrangement between c1 and c2

total = 6..whats OA?

bangalorian2000 · Answer

Possible patterns = 1x2x, 1xx2, x1x2, 2x1x, x2x1, 2xx1 = 6 hence D

Whats the OA

MyRedemption · Answer

Total cases : 12 ( student one has 4 options and student two has three options, 4x3=12)
Non-favourable cases : 6 (when two students sit together. students in desk 1 and desk 2 , in desk 2 and desk 3, in desk 3 and desk 4) for each of these cases there are two possibilities because the positions can be interchanged. hence 2x3=6.

SO favourable cases : 12-6=6.

TooLong150 · Answer

It is important to see that the empty desks are not distinct (I missed this the first time, because I considered them to be)
We have "4" students (2 normal students { A , B } and 2 students { E_1 , E_2 } that we can consider empty desks) and 4 places.
Total =  \frac{4!}{2!}  since  AE_1E_2B=AE_2E_1B 
Cases in which the two students are together {{ A , B },  E_1 ,  E_2 } = 3!*2!/2! since  ABE_1E_2  =  ABE_2E_1 
Total - together = 12 - 6 = 6

EMPOWERgmatRichC · Answer

Hi All,

Since the answers to this question are all small, and the question has a strong 'visual' component to it, we can 'brute force' the answer by drawing some pictures.

If we call the two students 'X' and 'Y', we have the following possibilities:

X _ Y _
Y _ X _

X _ _ Y
Y _ _ X

_ X _ Y
_ Y _ X

Total options: 6

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

armaankumar · Answer

Students: A, B
Desks: _ _ _ _
Case 1: A _ B _ (A&B can switch places) : 2 Ways
Case 2: _ A _ B ( A&B can switch places) : 2 Ways
Case 3: A _ _ B (A &b B can switch places): 2 Ways

Total Ways: 2 + 2 + 2 = 6

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How many different ways can 2 students be seated in a row of

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How many different ways can 2 students be seated in a row of

Prep Toolkit

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