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

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.
_________________

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.
_________________

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.
_________________

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.
_________________

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

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________