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# A parking lot has 16 spaces in a row. Twelve cars arrive, each of whic

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Senior Manager
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A parking lot has 16 spaces in a row. Twelve cars arrive, each of whic  [#permalink]

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20 Mar 2019, 05:59
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Difficulty:

95% (hard)

Question Stats:

27% (02:58) correct 73% (03:20) wrong based on 30 sessions

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A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

A) $$\frac{11}{20}$$

(B) $$\frac{4}{7}$$

(C) $$\frac{81}{140}$$

(D) $$\frac{3}{5}$$

(E) $$\frac{17}{28}$$
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Posts: 7958
Re: A parking lot has 16 spaces in a row. Twelve cars arrive, each of whic  [#permalink]

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20 Mar 2019, 06:21
1
1
A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

A) $$\frac{11}{20}$$

(B) $$\frac{4}{7}$$

(C) $$\frac{81}{140}$$

(D) $$\frac{3}{5}$$

(E) $$\frac{17}{28}$$

OK... We have 16 spaces and 12 occupy spaces, so the 4 vacant can be in 16C4 ways.
Now, it is easier to find combinations or cases when the 4 vacant are not next to each other.
For this, place 12 SUVs first, so there will be 13 places for these 4 vacant places, so 13C4. => _1_2_3_4_5_6_7_8_9_10_11_12_

So, Probability that the vacant places are available = $$\frac{16C4-13C4}{16C4}=\frac{16*15*14*13-13*12*11*10}{16*15*14*13}=\frac{13*12*10(28-11)}{28}=\frac{17}{28}$$

E
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A parking lot has 16 spaces in a row. Twelve cars arrive, each of whic  [#permalink]

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22 Mar 2019, 05:47
can you please let me why my logic is wrong :
total number of ways : 16C4
now let's count available ways for 2 adjacant spaces :
let a :2 adjacent places as a unit , b : one space , c : a car
now we have abbccccccccc ,let's arrange them in 15!/(12!*2!)
we divided by 12! because c is repeated , the same for 2! ( b is reapeted )
thus p = {15!/(12!*2!)} / 16C4
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Re: A parking lot has 16 spaces in a row. Twelve cars arrive, each of whic  [#permalink]

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26 Aug 2019, 12:54
chetan2u wrote:
A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

A) $$\frac{11}{20}$$

(B) $$\frac{4}{7}$$

(C) $$\frac{81}{140}$$

(D) $$\frac{3}{5}$$

(E) $$\frac{17}{28}$$

OK... We have 16 spaces and 12 occupy spaces, so the 4 vacant can be in 16C4 ways.
Now, it is easier to find combinations or cases when the 4 vacant are not next to each other.
For this, place 12 SUVs first, so there will be 13 places for these 4 vacant places, so 13C4. => _1_2_3_4_5_6_7_8_9_10_11_12_

So, Probability that the vacant places are available = $$\frac{16C4-13C4}{16C4}=\frac{16*15*14*13-13*12*11*10}{16*15*14*13}=\frac{13*12*10(28-11)}{28}=\frac{17}{28}$$

E

Sir,

Could you please elaborate the 13C4 part. This question nailed me to the cross!
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Re: A parking lot has 16 spaces in a row. Twelve cars arrive, each of whic  [#permalink]

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26 Aug 2019, 12:56
A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

A) $$\frac{11}{20}$$

(B) $$\frac{4}{7}$$

(C) $$\frac{81}{140}$$

(D) $$\frac{3}{5}$$

(E) $$\frac{17}{28}$$

Could you post the official solution and source??
i was totally taken for a ride
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Re: A parking lot has 16 spaces in a row. Twelve cars arrive, each of whic  [#permalink]

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26 Aug 2019, 14:55
A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers choose their spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

A) $$\frac{11}{20}$$

(B) $$\frac{4}{7}$$

(C) $$\frac{81}{140}$$

(D) $$\frac{3}{5}$$

(E) $$\frac{17}{28}$$

Warning: this is an "800++" level problem. You are VERY unlikely to see anything this difficult on the actual GMAT - it requires a sort of thinking about combinatorics that is virtually never useful on this test. I would strongly recommend against spending study time working through this solution, unless you're either already scoring at the 49+ level on Quant (and probably not even then), or you have some personal curiosity about this one.

The total number of ways that the 12 cars can park is (16 choose 12). The harder part is finding the number of scenarios in which Em is able to park. Let's focus on that.

Let's designate three types of parking spots: full spots, empty spots, and Em's Spot. We want to arrange 12 full spots, 2 empty (non-Em) spots, and Em's spot, in some kind of order. How many ways can we do that? Here's one example, where F = full, X = empty, and E = Em:

F X F X F F F F F F F F F E F

Note that because we're counting Em's spot as if it's one spot, our effective number of spots is reduced by 1, to 15. So, since we're choosing 12, 2, and 1 out of 15, the total number of arrangements is 15!/(12!*2!*1!) = (15 * 14 * 13)/2 = 15*13*7.

However, it's not that easy. We double-counted some possibilities. For instance, we counted this arrangement:

F X X F E F F F F F F F F F F

and we also counted this arrangement separately:

F E F X X F F F F F F F F F F

But, if you're just looking at parking spots, those are actually the exact same arrangement: spots 2 and 3 are empty, and spots 5 and 6 are empty. We only want to count that arrangement once, not twice! We're going to have to subtract out some cases where we double or triple counted.

First of all, if all four of the empty spots are next to each other, we accidentally counted that case 3 times, but we only want to count it once. There are 13 possible sets of 4 empty spots adjacent to each other, so we need to subtract 2*13.

Next, if three of the empty spots are next to each other, but the other one is somewhere else, we accidentally counted that case 2 times, but we only want to count it once. How many times did that happen? It might have happened in this situation:

X X X F (plus 12 spots with 1 empty)

Or it might have happened in this situation (or in the mirror image of this situation):

(some number of full spots) F X X X F (some number of full spots with 1 empty)

Basically, if the three empty spots in a row were on the end, there are 12 ways for the remaining cars to be arranged. If the three empty spots in a row were somewhere else, there are 11 ways for the remaining cars to be arranged. So, there are 2(12) + 12(11) possible ways to have three empty spots in a row, and we need to subtract 2(12)+12(11).

Finally, if there are two pairs of two empty spots each, and those two pairs aren't adjacent, we counted that case twice: once where Em parked in the first pair of spots, and once where she parked in the second pair of spots.

How many ways can we have two non-adjacent pairs of empty spots? I won't show the math here (ask if you can't work it out!) but the total comes out to 12 + 11 + 10 + 9 + ... + 2 + 1 = 13*6 possible ways. Subtract 13*6.

We're ready to do the final calculation. PHEW! The total number of possible arrangements where Em can park is 15*13*7 - 2*13 - 2*12 - 11*12 - 6*13. Simplify this:

15*13*7 - 2*13 - 12*13 - 6*13

15*13*7 - 20*13

(15*7-20)*13

85*13

There are 85*13 ways for Em to park, out of a total of (16 choose 12) arrangements. The probability is:

85*13 / (16! / (12!*4!)) = 85*13 / (16*15*14*13 / (4*3*2))

= 85 * 13 * 4 * 3 * 2 / (16*15*14*13)

= 85 * 4 * 3 * 2 / (16 * 15 * 14)

= 17 * 4 * 3 * 2 / (16 * 3 * 14)
= 17 * 2 / (4 * 14)
= 17 / (2 * 14)
= 17 / 28

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Re: A parking lot has 16 spaces in a row. Twelve cars arrive, each of whic   [#permalink] 26 Aug 2019, 14:55
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