A repair shop has six cars awaiting service: two hatchbacks,

Question

A repair shop has six cars awaiting service: two hatchbacks, two sedans, and two SUVs. The cars are serviced in a random order. What is the probability that the two hatchbacks are serviced consecutively and the two sedans as well?

(A) \(\frac{1}{30}\)
(B) \(\frac{1}{15}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{2}{15}\)
(E) \(\frac{2}{9}\)

(A)  \frac{1}{30} 
(B)  \frac{1}{15} 
(C)  \frac{1}{9} 
(D)  \frac{2}{15} 
(E)  \frac{2}{9}

paragw · Answer

Here's my solution: https://gmatclub.com/forum/download/file.php?id=88857 GMAT-Club-Forum-41akcj0y.png

sarkarsarbani · Answer

In place of 4!,won't it be 2! since 2H and 2S are considered as 2 blocks and not 4 units?

egmat · Answer

Hi sarkarsarbani,

Good instinct to question the count, but let's recount the  units  being arranged.

When we glue the hatchbacks into one block and the sedans into another, those two blocks are not the  only  things left to order. The  two SUVs are still loose  - they never got glued to anything. So the full list of items to arrange is:

-  
-  
- SUV1
- SUV2

That's  4  separate units, not  2  - which is exactly why it's  4!  =  24 , not 2!.

The trap here is picturing only the glued pairs and forgetting the leftover cars that travel on their own. A quick check: the units have to cover all  6  cars. Two blocks of  2  cars account for only  4  cars; the remaining  2  SUVs must show up as their own units, giving  4  units in total.

From there the rest of the posted solution holds:  4!  to arrange the units, times  2!  inside the hatchback block, times  2!  inside the sedan block -  24 × 2 × 2 = 96  favorable, over  6! = 720  total =  2/15 .

Lock it in with a smaller version:  suppose you only had H, H, X ( 1  pair plus one loose car) and wanted the two H's together. Glue the H's into a block - now you arrange   and X, which is  2  units -  2! . The loose car always counts as its own unit. Scale that up and you'll always catch every leftover item before choosing the factorial.

Answer: D

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=88866 GMAT-Club-Forum-8s35cfz5.png

vasu1104 · Answer

we have 6 cars.
we need to find that two hatchbacks are serviced consecutively and two sedans are as well.

step 1- total ways= 6! = 720

step 2- now since two hatchbacks and sedans are serviced consecutively independent to each other. lets consider two hatchbacks as one and two sedans as one.

so now we have 4 cars only. so ways to arrange them = 4!.
but here the 2 hatchbacks can be arranged in 2! and two sedans can be arranged in 2!

so in total= 4!*2!*2! = 96

so probability = 96/720 = 2/15

choice D

A repair shop has six cars awaiting service: two hatchbacks,

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A repair shop has six cars awaiting service: two hatchbacks,

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