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# Josh has to run an electrical wire from point a to point b

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Josh has to run an electrical wire from point a to point b  [#permalink]

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Updated on: 30 Jun 2013, 13:27
4
22
00:00

Difficulty:

25% (medium)

Question Stats:

65% (00:58) correct 35% (01:36) wrong based on 502 sessions

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Josh has to run an electrical wire from point a to point b along a circuit that is restricted to the grid shown to the left. How many possible paths could Josh use that have the minimum possible length?

A. 8
B. 10
C. 12
D. 15
E. 16

Originally posted by zisis on 28 Aug 2010, 08:58.
Last edited by Bunuel on 30 Jun 2013, 13:27, edited 2 times in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 50583
Re: Grockit: similar to OG Quant qustion  [#permalink]

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28 Aug 2010, 09:05
12
7
zisis wrote:
Josh has to run an electrical wire from point a to point b along a circuit that is restricted to the grid shown to the left. How many possible paths could Josh use that have the minimum possible length?

A 8
B 10
C 12
D 15
E 16

obv the answer can be found by counting the routes, but is there a better way?

You can notice that in order the length to be minimum wire should only go UP and RIGHT: namely twice UP and 4 times RIGHT.

So combination of UURRRR: # of permutations of 6 letters out of which there are 2 identical U's and 4 identical R's is $$\frac{6!}{2!4!}=15$$.

Hope it's clear.
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Re: Grockit: similar to OG Quant qustion  [#permalink]

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28 Aug 2010, 09:34
2
Excellent approach.

Bunnel I was just wondering how much you got on Gmat? Have you given?
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Re: Grockit: similar to OG Quant qustion  [#permalink]

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29 Aug 2010, 04:48
Bunuel wrote:
zisis wrote:
Josh has to run an electrical wire from point a to point b along a circuit that is restricted to the grid shown to the left. How many possible paths could Josh use that have the minimum possible length?

A 8
B 10
C 12
D 15
E 16

obv the answer can be found by counting the routes, but is there a better way?

You can notice that in order the length to be minimum wire should only go UP and RIGHT: namely twice UP and 4 times RIGHT.

So combination of UURRRR: # of permutations of 6 letters out of which there are 2 identical U's and 4 identical R's is $$\frac{6!}{2!4!}=15$$.

Hope it's clear.

thanks! thats exactly what i was looking for ! If i recall correcty, you must be the GMATclub combinations expert

[ IDEA how about we have experts stamps for certain individuals ! SC, combinations, RC, algebra etc....something the forum admins should consider...]
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Re: Grockit: similar to OG Quant qustion  [#permalink]

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29 Aug 2010, 08:04
He is expert in almost every topic of quant.
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Re: Grockit: similar to OG Quant qustion  [#permalink]

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29 Aug 2010, 09:18
lol...then he will have lots of "stamps" :p
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find the no of ways  [#permalink]

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23 Oct 2010, 05:48
hi all
if there is a rectangle and this rectangle is divided in to four equal rectangles by building roads inside it then how many ways are there with which one can reach from one corner to other diagonally opposite corner.
please try to explain in terms of combinations .like in terms of nCr etc.

thanks
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Re: find the no of ways  [#permalink]

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23 Oct 2010, 09:11
harshsingla wrote:
hi all
if there is a rectangle and this rectangle is divided in to four equal rectangles by building roads inside it then how many ways are there with which one can reach from one corner to other diagonally opposite corner.
please try to explain in terms of combinations .like in terms of nCr etc.

thanks

You need to find permutations of UURR = 4!/2!2! = 6
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Re: find the no of ways  [#permalink]

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23 Oct 2010, 09:21
but how 4c2..???
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Re: find the no of ways  [#permalink]

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23 Oct 2010, 09:27
1
harshsingla wrote:
but how 4c2..???

think of it like this
You have four moves to make to reach the opposite corner

two moves are up
and two moves are right

now, any order of moves would always get to the corner

the question really is that out of move1,2,3&4 which two you pick to be the Up move (the other two will be the right move). The ways to do this is C(4,2), choosing 2 out of 4.
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Re: Josh has to run an electrical wire from point a to point b  [#permalink]

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09 Jun 2016, 00:14
zisis wrote:
Attachment:
picture.jpg
Josh has to run an electrical wire from point a to point b along a circuit that is restricted to the grid shown to the left. How many possible paths could Josh use that have the minimum possible length?

A. 8
B. 10
C. 12
D. 15
E. 16

out of six we need to choose 4 on x-axis or out of six we need to choose 2 on y axis
so 6c2 or 6c4
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Re: Josh has to run an electrical wire from point a to point b  [#permalink]

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27 Jun 2017, 10:20
1
Top Contributor
1
zisis wrote:
Attachment:
picture.jpg
Josh has to run an electrical wire from point a to point b along a circuit that is restricted to the grid shown to the left. How many possible paths could Josh use that have the minimum possible length?

A. 8
B. 10
C. 12
D. 15
E. 16

This is a great candidate for applying the Mississippi Rule
The rule is useful for arranging a group of items in which some of the items are identical.
It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

-----NOW ONTO THE QUESTION--------
To get from point a to point b, we must travel UP (U) two times, and travel RIGHT (R) 4 times
In other words, we want to determine the number of DIFFERENT ways to arrange 2 U's and 4 R's
let's apply the above rule.

There are 6 letters in total
There are 2 identical U's
There are 4 identical R's
Total number of possible arrangements = 6!/[(2!)(4!)]
= 15

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Re: Josh has to run an electrical wire from point a to point b  [#permalink]

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27 Jul 2018, 18:33
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Re: Josh has to run an electrical wire from point a to point b &nbs [#permalink] 27 Jul 2018, 18:33
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