Question

How many different signals that can be given using any number of flags from 5 flags of different colors is?

Answer 1

@ganeshie8

Answer 2

@ParthKohli

Answer 3

Wow, I totally hate these questions. :(
Just make cases?

Answer 4

idk the options are 325,525,425,625

Answer 5

\[\sum\limits_{n=0}^{5} \dbinom{5}{n}*n! \]

should do right ?

Answer 6

\[\binom{5}1\cdot 1! + \binom{5}2 \cdot 2! + \binom{5}3 \cdot 3! + \binom{5}4\cdot 4! + \binom{5}5\cdot 5!\]Almost like ganeshie's answer except \(n = 0\) is not there

Answer 7

yeah treating "no flags" as a signal makes no sense

Answer 8

actually any any number of flags is what is bugging me
and same color are identical

Answer 9

I know, that language is confusing. It just means that you don't have to use all the flags to make a signal.

Answer 10

so have to use only 5 at max?

Answer 11

It's like forming power set of a given set. In our case order matters

Answer 12

Suppose I give you a set with 5 elements. How many subsets can you make ?

Answer 13

2^5-1

Answer 14

empty set is also a subset
so 2^5 subsets in total, right ?

Answer 15

ok..

Answer 16

Can you show me how you got 2^5 ?

Answer 17

yep
element 1 will have 2 choices whether to go or not
same with rest

Answer 18

That is one nice way to look at it.

But it is useless here

Answer 19

A more useful derivation of the same result is by considering different cases :

1) number of subsets with 0 elements= ?
2) number of subsets with 1 elements= ?
3) number of subsets with 2 elements= ?
4) number of subsets with 3 elements= ?
5) number of subsets with 4 elements= ?
6) number of subsets with 5 elements= ?

Answer 20

nC0 + n C 1 ...... nC5

Answer 21

Yep!
How to make that logic useful in our problem ?

Answer 22

if it helps
5P1 + 5P2 + 5P3 + 5P4 + 5P5

Answer 23

i get it..actually misunderstood the language
thanks guys :)

Answer 24

there one or two more
I was trying to converge this series
(2r+1) C r + (2r+1) C r-1 ........ (2r+1) C 1

Answer 25

Oh, good. Add (2r+1)C0 then try to think what the relation of that sum and...

(2r+1)C0 + (2r+1)C1 + ... + (2r+1)C(2r+1) is

Answer 26

You could also stare at pascal triangle a few seconds to get a feel of that sum...

Answer 27

but the series is starting from 2r+1C r

Answer 28

i never actually understood what the pascal triangle is for

Answer 29

Oh let me give you a brief introduction first then

Answer 30

cool :)

Answer 31

Pascal triangle starts from the top row with 1 :

Answer 32

Next row is created from the present row by doing below :
1) append 1's at left and right.
2) add two adjacent numbers in present row and put the result in the next row

Answer 33

is that because\[\binom{n}{r-1}+\binom{n}r = \binom{n+1}r\]

Answer 34

yep