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

Question

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

Show Answer

Official Answer

D

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

Bunuel · Accepted Answer

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).

Answer: D.

rxs0005 · Answer

Bunuel

which formula is this ?

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

can you explain that

thanks

Bunuel · Answer

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.

rxs0005 · Answer

Thanks , i was not aware of this formula

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

Bunuel · Answer

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.

rxs0005 · Answer

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

Bunuel · Answer

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.

gtr022001 · Answer

this is a great explanation on permutations, would it be possible to add it to the GMAT math book?

mvictor · Answer

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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