Question

Find How many arrangements of the letters of the word UNIFORM are possible if the M is somewhere to the right of the U?

Answer 1

Why can 7P5 give the correct answer? i understand that it is just half of 7!, but why does taking 5 letters out of 7 work?

Answer 2

@Hero what you say here...?

Answer 3

will it be 7P6...? or not...?

Answer 4

I would solve this using 6 different cases, in which we place the U, then place M in any position to the right of U, and any permutation of the remaining 5.

So, we have 7 positions. _ _ _ _ _ _ _

First, place U in position 0. U _ _ _ _ _ _

M can be anywhere to the right, so 6 positions, and the remaining 5 can be any arrangement for the remaining empty positions, 5!.

Second, place U in position 1. _ U _ _ _ _ _

M can be anywhere to the right, so 5 positions, and remaining have 5! positions.

So, here's the pattern. 6*5! + 5*5! + ... + 1*5!.

That is 5!(1+2+3+4+5+6) = 5!(21)

Answer 5

so 7P6 is rite...?

Answer 6

Well, our answers don't agree.

Answer 7

hmmm ok... well you take it combination or permutation...?

Answer 8

Order matters, so I'm counting different orders in my formulation.

Answer 9

ok... how many words are there in the given "word"...?

Answer 10

5!*21=2520, half of 7! not coincidentally.

Answer 11

there are 7 words in UNIFORM... agree...?

Answer 12

Yes. Take me through your argument. Do note, however, that this problem is not a simple permutation problem: we have a constraint. M must be to the right of U.

Answer 13

yeah.. .agree...

Answer 14

so how many possible places are there for "M"...?

Answer 15

The answer you gave, 7P6, is the same as 7!, as though there were no constraints.

Answer 16

The # of positions for M depends on where U is placed.

Answer 17

ahaan... rite... so U will have its own place....

Answer 18

Yes, my approach was to break it down into 6 different cases (U cannot be in the final position). Very quickly, a pattern emerged, and that's how I arrived at my solution.

Answer 19

@dg2 what you say...?

Answer 20

6 ways so 5!

Answer 21

I'm convinced my case-by-case analysis is correct. I posted it earlier. if you have any specific questions about it, I'd be happy to try to answer any of them.

Answer 22

why u are multiplying 5! to 5 ,4,3,2,1 i dont understand

Answer 23

it's factorial... thats why.... @dg2

Answer 24

5! = 5*4*3*2*1
10! = 10*9*8*7*6*5*4*3*2*1

Answer 25

There are 6 possible locations for U. For each location of U, I place the M somewhere to the right, and then the 5 remaining letters have 5! possible arrangements in the 5 remaining empty positions.

Answer 26

oh.no no i am not asking that @paki.

Answer 27

o, here's the pattern. 6*5! + 5*5! + ... + 1*5!. this thing i asked @queelius to clarify

Answer 28

oh i understood @queelius .u r right

Answer 29

Case 1: when U is in the first position, M can be in 6 different positions. The remaining letters have 5! arrangements.

Case 2: when U is in the second position, M can be in 5 different positions. The remaining letters have 5! arrangements.

...

Case 6: when U is in the 6th position, M can be in 1 position. The remaining letters have 5! arrangements.

Sum the cases up.

Answer 30

Okay.

Answer 31

nice:-).good

Answer 32

Turns out, in hindsight, it's clear why 7P5 works. I didn't see that except after I did my case by case analysis.

Answer 33

7p5 is the answer ? idk

Answer 34

5!*21=7p5

Answer 35

7p5 means

Answer 36

Out of 7 items, find every arrangement (permutation) of 5 of them.

7 permute 5.

Answer 37

7!/2! right?

Answer 38

Yes.

Answer 39

k .3q