IMHO A is the answ
if the answer is ok, I can explain my logic

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~

@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.

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

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.

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

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

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

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

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

Because we are ASSIGNING combinations of 3 to each courier, the ORDER of the groupings matters.
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.

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

My first answer was a straight C, over thought it and messed it up ..

Thanks Mitch and Pemdas

good one pemdas.

perhaps you should have been more wicked -- by adding an answer choice such as 21!/7!(3!)^7 :twisted:

Cheers,
LB

next time will be done, here my intention was to leave only the ordered arrangement
Cheers
pemdas

Good one. Thanks PEMDAS

hi Mike,
kindly explain as to why 3! is there in the denominator. i didnt understand.

GMATGuruNY wrote:

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

Because we are ASSIGNING combinations of 3 to each courier, the ORDER of the groupings matters.
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.