When 10 people in bank line up for deposit, Vivek insists on

Question

When 10 people in bank line up for deposit, Vivek insists on being somewhere ahead of bewakoof, If Vivek's demand is satisfied, in how many ways can the people line up?

gmatacer · Answer

bewakoof can be at any position but 1 because all time vivek is ahead of him. 
we can line up all people in 9! ways

ywilfred · Answer

It's true that Bewakook must not be at position 1 in the queue, but it's not correct to multiple 9! for all 9 times. If Bewakoof was at position 4, and you use 9!, then there will be some permutations that have Vivek behind Bewakoof because 9! considers all possible arrangements.

Here's what I think:

If B is in queue position 2, then V must be at 1 --> number of ways = 8!

If B in in queue position 3, and V at 2 --> 8! ways
If B is in queue position 3, and V at 1 --> 8! ways

If B is in queue position 4, and V at 3 --> 8! ways
If B is in queue position 4, and V at 2 --> 8! ways
If B is in queue position 4, and V at 1 --> 8! ways

So for B in queue position 2 to position 9, we will have 8!(2+3+4+5+6+7+8)

For B in last position there are 9! ways to arrange since V will always be in front.

Total number of way = 36(8!) + 9!

giddi77 · Answer

I have a vague solution, but it works for 3 and 4 people and I generalized it to 10.. 

Let's say there are 3 people: Vivek,Bewakoof,Ywilfred --V,W,Y
Total arrangements = 6
VBY
  BVY  
VYB
  BYV  
YVB
  YBV  

We cannot have the arrangements that are marked   RED  . In other words, they are repititions. VB is same as BV...
Hence we need to divide total arrangements by 2! to remove the repititions. Total arrangements = 3!/2! = 3

Similarly for 4 people it is 4!/2! = 12

Finally for 10 people it is 10!/2! = 1814400

jodeci · Answer

I think:

9(9*8!) because the pattern will occur nine times.

Basically, 9*9!

allabout · Answer

I think he meant it differently:

XXXXXXXXVB so Vivek had 9 positions to choose from
XXXXXXXVBY so Vivek had 8 positions to choose from
XXXXXXVBYY " 7

etc.

Each time Bewakoof places himself one position forward, Vivek can choose from one place less. Since Bewakoof is at another position in each case, we can speak of different combinations.

That's the way he comes at 9!, and he's correct so far. We have to consider the other people too, that's why we multiply with 8 finally.

BG · Answer

10 people can be arranged in 10! ways. In half of the outcomes B will be before V and in half of the arrangements V will be before B. So think that it should be 10!/2, IMO

trivikram · Answer

1+2+3+4+5+6+7+8+9 = 45 WAYS If Bewkoof (I really do not mean it :-D

) is in 10th position Viv has 9 places and so only 9 ways not 9!. We are not worried about others arrangements but only Vivek's Like the same way I got the above answer

vivek123 · Answer

Hmmm...

Let 10 people including myself (X) & Abhishek (Y) be:

A B C D E F G H X Y
Front -----------Back

1) Order XY = 9!
2) A B C D E F G X H Y = 8! * 8
3) A B C D E F X G H Y = (8*6) * 7! = 7! * 56
4) A B C D E X F G H Y = (8*7*6) * 6 !
5) A B C D X E F G H Y = (8*7*6*5) * 5!
so on...
& last one is
X A B C D E F G H Y = 8!

Does it make sense? Can anybody check this? :stupid

shevy · Answer

I agree, I think it is 9 * 9 * 8 !

Professor · Answer

buddy, OE is appreciated..................

BG · Answer

Hallo Prof,
I think that the explanation goes like this. 10 people can line in 10! ways. In half of the arrangements Vivek will be before bevakof and in the other half of the arrangements it will be other way round. That is how it comes to 10!/2

Professor · Answer

wow, that is simple and best........thanxxxxx :wink:

chiragr · Answer

I kind of Used same method but I got 8! *  = 45*8!

A on Position B different Position Other
last 9 8! = 1*9*8!
last-1 8 8! = 1*8*8!
.
.
.
.
.
.
Second 1 8! = 1*1*8!

chiragr · Answer

Yep this is the simple and perhaps the best way of solving...

uanant · Answer

Let me know if this makes sense
There are 10 People, if you put vivek and bewakoof as a bundle then the number of people and their positions reduces to 9. However vivek and bewakoof can not switch position among themselves so their order remains one.
So then number of way they can be arranged would be equal to 9!

devansh_god · Answer

hey conocieur!!!
thats exactly wot i thought when i first saw the question.......... donno if it is right or wrong with so many ppl with diff ans n explanations.....

conocieur · Answer

I couldn't explain it better than this, thanks BG

HongHu · Answer

This is a good thinking, but not correct. This method assumes that vivek and bewakoof will always stay together. However I suspect that they may not have to be that close.

The 10!/2 is the best and shortest approach to this, in my opinion.

uanant · Answer

I am sorry i realized my mistake how about this approach.
Total number of ways they can be arranged = 10!
Always together = 9!
Never together = 10! - 9!
in never together case then can switch positions among themselves which would be = 2. If we fix their order of apparance then never together event will reduce to half.
Therefore total number of way based on given condition =
 
 
 
9!*5.5

When 10 people in bank line up for deposit, Vivek insists on

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