Gita, Hussain, Inge, Jeong, Karen, and Leila are seated in a row of

Question

Gita, Hussain, Inge, Jeong, Karen, and Leila are seated in a row of six chairs. How many seating arrangements are possible if Gita cannot sit next to Inge and Jeong must sit next to Leila?

(A) 288
(B) 240
(C) 144
(D) 120
(E) 96

KarishmaB · Accepted Answer

This is correct. There are 240 ways in which you can make the 6 people sit together such that J and L are together. Now, forget it.

Think about the number of ways in which you can make I and G sit together while J and L are sitting together:

J and L are tied together and now we tie I and G together. So we have 4 individuals/pairs. We can arrange them in 4! ways but J and L can switch places and so can I and G. So total number of ways = 4! * 2 * 2 = 96

There are 96 ways in which I and G will be together and J and L will also be together.

So in 240 - 96 = 144 ways I and G will not be together while J and L will be together.

zerosleep · Answer

I am taking he initials of each for refernce... Since, J and L have to be together, lets combine them into 1. So, now we have 5 things which can be arranged in 5! ways. ALso, J and L can be arranged in two ways (JL, LJ) . 
This gives total ways, 5! * 2 = 240

Now, we are given G and I can not be together, therefore subtract the cases when they (G and I) are together. Now, see we have 4 people (JL and GI) are comined as one. Total no of arrangement = 4!
But, J and L can be arranged internally in 2 ways (JL and LJ). Similarly, G and I can be internally arranged in two ways (GI and IG).
Therefore, total such cases will be 4! * 2 * 2 = 96. Subtract this from 240.
240 - 96 = 144. Hence, C

usre123 · Answer

I understand how we first band together J and L, and get 5! = 120. Then because J and L can switch places, we have 5! * 2= 240.

Now for I and G not sitting together, let's calculate for when they do sit together ( and assuming J and L are already sitting together):
 4! =24
So now I and G can switch, so 24 x 2= 48. 
So my final answer is 240- 48 =196.
I don't understand why we must accommodate for J and L switching, as I've already done that in the first part.

Thanks.

ScottTargetTestPrep · Answer

Let’s first assume that the first requirement (Gita cannot sit next to Inge) does not exist but the second (Jeong must sit next to Leila) does. Then we can assume Jeong and Leila as “1 person.” So the number of ways to arrange “5 people” is 5! = 120. However, within that “1 person” the seating arrangement of Jeong and Leila can be either JL or LJ. Therefore we have to multiply 120 by 2 to obtain 240 arrangements.

Of course, the 240 arrangements we’ve obtained above omits the first requirement. So now let’s consider that requirement. However, more precisely, let’s consider the opposite of that requirement; that is, let’s consider that Gita must sit next to Inge. If that is the case, like before, we can consider Gita and Inge as “1 person.” So the number of ways to arrange “4 people” is 4! = 24. However, these “4 people” consist of two “1 person” entities: Gita and Inge, and Jeong and Leila. Therefore we have to multiply 24 by 2 (for GI and IG) and again by 2 (for JL and LJ) to obtain 96 arrangements.

Since we are really looking for the number of arrangements where Gita and Inge cannot sit next to each other, then the number of such arrangements is 240 - 96 = 144.

Answer: C

Aboyhasnoname · Answer

How I did....

Number of Ways in which J & L are together is 5! X 2!

In this G and I are together also and not together also....From this remove the ones that they are together.....

Consider this....... _ _ (_ _ ) (_ _)
4! x 2! x 2! ...

5! X 2! - 4! x 2! x 2! = 144

Gita, Hussain, Inge, Jeong, Karen, and Leila are seated in a row of

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