In how many ways can 21 post mails be distributed among seven couriers such that each courier receives an equal amount of post mails?
A 21^7
B (7!)^7
C 21!/(3!)^7
D 21!/7!
E 7^21
made up
post mails
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Assuming all 21 mails are distinct... We need to distribute them in 7 groups of 3.
21 different posts can be arranged in 21! ways..
Now since each courier receives 3 posts each, the order of the 3 doesnt matter..
For the 1st courier, we have (21*20*19)/3!
For the 2nd courier, we have (18*17*16)/3! and so on..
Combing we have 21!/(3!^7) .. C
~Updated~
21 different posts can be arranged in 21! ways..
Now since each courier receives 3 posts each, the order of the 3 doesnt matter..
For the 1st courier, we have (21*20*19)/3!
For the 2nd courier, we have (18*17*16)/3! and so on..
Combing we have 21!/(3!^7) .. C
~Updated~
Last edited by shankar.ashwin on Fri Feb 03, 2012 11:25 am, edited 2 times in total.
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I am not sure of the formula... But this is similar to a problem in GMATPrep.. discussed several times here..
I think Mitch explains it well.. Have a look
https://www.beatthegmat.com/forming-teams-t73034.html
I think Mitch explains it well.. Have a look
https://www.beatthegmat.com/forming-teams-t73034.html
LalaB wrote:@shankar.ashwin
do u mean we need to use this formula?
The number of ways in which mn different items can be divided equally into m groups is (mn)!/(n!)^m.
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nice, i made this up not for enjoyment and easy answers :mrgreen:
am i nasty? :twisted:
@Shankar, yes Mitch has solved and briefly explained, but did you get there?
it's still pending solution; i have already made up another "nice" question & will post after you hit this one
am i nasty? :twisted:
@Shankar, yes Mitch has solved and briefly explained, but did you get there?
it's still pending solution; i have already made up another "nice" question & will post after you hit this one
Success doesn't come overnight!
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Pemdas.. Just edited my post... Still dont get to any of the answer choices mentioned... Not sure whats wrong with the solution..
pemdas wrote:nice, i made this up not for enjoyment and easy answers :mrgreen:
am i nasty? :twisted:
@Shankar, yes Mitch has solved and briefly explained, but did you get there?
it's still pending solution; i have already made up another "nice" question & will post after you hit this one
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yea, similar
updated the thread with my solution about the same https://www.beatthegmat.com/forming-team ... tml#448600 . Here -hint, Order MATTERS! hit it please
updated the thread with my solution about the same https://www.beatthegmat.com/forming-team ... tml#448600 . Here -hint, Order MATTERS! hit it please
shankar.ashwin wrote:Pemdas.. Just edited my post... Still dont get to any of the answer choices mentioned... Not sure whats wrong with the solution..
pemdas wrote:nice, i made this up not for enjoyment and easy answers :mrgreen:
am i nasty? :twisted:
@Shankar, yes Mitch has solved and briefly explained, but did you get there?
it's still pending solution; i have already made up another "nice" question & will post after you hit this one
Success doesn't come overnight!
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Because we are ASSIGNING combinations of 3 to each courier, the ORDER of the groupings matters.pemdas wrote:In how many ways can 21 post mails be distributed among seven couriers such that each courier receives an equal amount of post mails?
A 21^7
B (7!)^7
C 21!/(3!)^7
D 21!/7!
E 7^21
made up
Assigning ABC to courier 1 and DEF to courier 2 is not the same as assigning DEF to courier 1 and ABC to courier 2.
Step 1: Count how many combinations of 3 can be assigned to each courier.
Step 2: Multiply the results in order to count the number of ways in which these combinations of 3 can be ORDERED among all 7 couriers.
Combinations of 3 that could be assigned to the first courier:
(21*20*19)/3!.
Combinations of 3 that could be assigned to the second courier:
(18*17*16)/3!.
Combinations of 3 that could be assigned to the third courier:
(15*14*13)/3!.
And so on.
When we multiply these results:
The numerator will be 21!: (21*20*19)*(18*17*16)*(15*14*13)....
Since each of the 7 groupings is divided by 3!, the denominator will be (3!)(3!)(3!)(3!)(3!)(3!)(3!) = (3!)�.
Thus, the correct answer here is C: 21!/(3!)�.
This expression represents the number of ways to ARRANGE 7 combinations of 3.
In other words, the number of ways we can ASSIGN 3 letters to each of the 7 couriers.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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My first answer was a straight C, over thought it and messed it up ..pemdas wrote:yea, similar
updated the thread with my solution about the same https://www.beatthegmat.com/forming-team ... tml#448600 . Here -hint, Order MATTERS! hit it please
Thanks Mitch and Pemdas
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hi Mike,
kindly explain as to why 3! is there in the denominator. i didnt understand.
kindly explain as to why 3! is there in the denominator. i didnt understand.
GMATGuruNY wrote:Because we are ASSIGNING combinations of 3 to each courier, the ORDER of the groupings matters.pemdas wrote:In how many ways can 21 post mails be distributed among seven couriers such that each courier receives an equal amount of post mails?
A 21^7
B (7!)^7
C 21!/(3!)^7
D 21!/7!
E 7^21
made up
Assigning ABC to courier 1 and DEF to courier 2 is not the same as assigning DEF to courier 1 and ABC to courier 2.
Step 1: Count how many combinations of 3 can be assigned to each courier.
Step 2: Multiply the results in order to count the number of ways in which these combinations of 3 can be ORDERED among all 7 couriers.
Combinations of 3 that could be assigned to the first courier:
(21*20*19)/3!.
Combinations of 3 that could be assigned to the second courier:
(18*17*16)/3!.
Combinations of 3 that could be assigned to the third courier:
(15*14*13)/3!.
And so on.
When we multiply these results:
The numerator will be 21!: (21*20*19)*(18*17*16)*(15*14*13)....
Since each of the 7 groupings is divided by 3!, the denominator will be (3!)(3!)(3!)(3!)(3!)(3!)(3!) = (3!)�.
Thus, the correct answer here is C: 21!/(3!)�.
This expression represents the number of ways to ARRANGE 7 combinations of 3.
In other words, the number of ways we can ASSIGN 3 letters to each of the 7 couriers.