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# A team has to design a flag. The team has six blue strips of

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Director
Joined: 07 Jun 2004
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A team has to design a flag. The team has six blue strips of  [#permalink]

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22 Dec 2010, 08:21
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7
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:56) correct 42% (02:06) wrong based on 167 sessions

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A team has to design a flag. The team has six blue strips of cloth and seven red strips of cloth that they must use as is to make the flag. How many different flags can the team design with the materials at hand?

(A) 42
(B) 120
(C) 720
(D) 1,716
(E) 5,040

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Joined: 02 Sep 2009
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22 Dec 2010, 08:33
2
2
rxs0005 wrote:
A team has to design a flag . The team has six blue strips of cloth and seven red strips of cloth that they must use as is to make the flag. How many different flags can the team design with the materials at hand?

(A) 42
(B) 120
(C) 720
(D) 1,716
(E) 5,040

The # of permutations of total 6+7=13 strips where 6 are identical blue and 7 are identical red is 13!/(6!*7!)=1716 (note that they must use all the materials at hand, also I guess that the strips must be either only vertical or only horizontal).

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22 Dec 2010, 08:37
Bunuel

which formula is this ?

why do you divide 13! by 7! and 6!

can you explain that

thanks
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22 Dec 2010, 08:41
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rxs0005 wrote:
Bunuel

which formula is this ?

why do you divide 13! by 7! and 6!

can you explain that

thanks

Sure.

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:

$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.

For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

So, number of permutation of 13 strips out of which 7 are blue and 6 are red would be $$\frac{13!}{7!6!}$$.

Hope it's clear.
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22 Dec 2010, 08:48
Thanks , i was not aware of this formula

also here does the order matter ? ( i am assuming yes because its permutation and not combination )
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22 Dec 2010, 09:01
rxs0005 wrote:
Thanks , i was not aware of this formula

also here does the order matter ? ( i am assuming yes because its permutation and not combination )

Yes, order matters. Consider there to be 3 strips 2 red and 1 blue, there will be 3!/2!=3 different arrangements possible: RRB, RBR, BRR.

Also note that generally:
The words "Permutation" and "Arrangement" are synonymous and can be used interchangeably (order matters).
The words "Combination" and "Selection" are synonymous and can be used interchangeably.

Hope it's clear.
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22 Dec 2010, 10:38
Thanks for the de-breifing I read these books and they have this
n things can be arranged n! ways

eg :

ABA can be arranged 3! = 6 ways -- 1

so ABA can be also arranged in 3!/2! = 3 ways -- 2

what is the diff between 1 and 2

i am guessing 1 is wrong it cant be 6 ways ?

just trying to get the basic clear here

thanks
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22 Dec 2010, 10:58
3
rxs0005 wrote:
Thanks for the de-breifing I read these books and they have this
n things can be arranged n! ways

eg :

ABA can be arranged 3! = 6 ways -- 1

so ABA can be also arranged in 3!/2! = 3 ways -- 2

what is the diff between 1 and 2

i am guessing 1 is wrong it cant be 6 ways ?

just trying to get the basic clear here

thanks

Permutations (arrangements) of $$n$$ distinct things is $$n!$$. So, 3 distinct letters A, B, C can be arranged in 3!=6 ways: ABC, ACB, BAC, BCA, CAB, and CBA;

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is: $$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$. So, 3 letters AAB, of which 2 are alike, can be arranged in 3!/2!=3 ways: AAB, ABA, and BAA. That's because arrangements of 2 A's don't give different group or in another way AB can be arranged as AB or BA but AA can only be arranged in one way AA.

Hope it's clear.
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23 Dec 2010, 13:19
Bunuel wrote:

Sure.

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:

$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.

For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

So, number of permutation of 13 strips out of which 7 are blue and 6 are red would be $$\frac{13!}{7!6!}$$.

Hope it's clear.

this is a great explanation on permutations, would it be possible to add it to the GMAT math book?
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A team has to design a flag. The team has six blue strips of  [#permalink]

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08 Feb 2016, 19:23
oh damn, I hate this kind of problems..
first...we have 6 identical blues and 7 identical reds.
now, we have 13 totals.
thus, we can arrange in 13! ways.
since we have 6 blue and 7 red, we need to divide 13! by 6! and by 7!.
we get:
13*12*11*10*9*8*7!/7!*1*2*3*4*5*6
simplify:
13*12*11 = 1,726.

D
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Re: A team has to design a flag. The team has six blue strips of  [#permalink]

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