Dtjytyk

Where does the bonus feat in the cleric starting package come from?

Electoral considerations aside, what are potential benefits, for the US, of policy changes proposed by the tweet recognizing Golan annexation?

Why do we read the Megillah by night and by day?

Is there a single word describing earning money through any means?

How much character growth crosses the line into breaking the character

Problem with TransformedDistribution

Is this toilet slogan correct usage of the English language?

Request info on 12/48v PSU

How could a planet have erratic days?

Count the occurrence of each unique word in the file

How can "mimic phobia" be cured or prevented?

Why should universal income be universal?

Delivering sarcasm

Why does the Sun have different day lengths, but not the gas giants?

What should you do when eye contact makes your subordinate uncomfortable?

How to implement a feedback to keep the DC gain at zero for this conceptual passive filter?

Are the IPv6 address space and IPv4 address space completely disjoint?

Calculating Wattage for Resistor in High Frequency Application?

What was this official D&D 3.5e Lovecraft-flavored rulebook?

Why did the EU agree to delay the Brexit deadline?

When a Cleric spontaneously casts a Cure Light Wounds spell, will a Pearl of Power recover the original spell or Cure Light Wounds?

GraphicsGrid with a Label for each Column and Row

How can we generalize the fact of finite dimensional vector space to an infinte dimensional case?

The screen of my macbook suddenly broken down how can I do to recover

Simplify $P=1-left( 364 times 363 cdots cdots frac{364-N+1}{364^N} right)$

$p_n = 1- left( 1-frac{1}{365}right)left( 1-frac{2}{365}right)cdots left( 1-frac{n-1}{365}right)$ then $p_n>frac{1}{2}$ for $n>?$Calculating $left ( frac{-3}{p} right ) $ when $p=u^2+3v^2$Find $n$ such that the numerator of $1+frac{1}2+frac{1}3+cdots+frac{1}n$ is a square numberHow prove this $1times 3times 5times 7cdotstimes 2009+2times4times6cdots 2010equiv 0 (mod2011)$$left[frac n1right]+ left[frac n2right] + cdots+left[frac nnright]+left[sqrt nright]$ is evenShow that $left[alpharight]+left[alpha+frac{1}{m}right]+cdots + left[alpha+frac{m-1}{m}right] = left[malpharight]$Evaluate $left(frac{3}{11}right)$ using Euler's Criterion…Solving birthday problem without complementShow that $left(frac{1+sqrt{5}}{2}right)^5 > 10$Number of integers satisfying $left[frac{x}{100}left[frac{x}{100}right]right]=5$

0

$begingroup$

Show that after simplification of $P=1-left( 364 times 363 cdots cdots frac{364-N+1}{364^N} right)$, we get the result $$P=1-e^{-frac{N(N-1)}{2 *364}}. $$

Answer:

We have,

$P=1-left( 364 times 363 cdots cdots frac{364-N+1}{364^N} right) \ Rightarrow P=1-frac{364 times 363 cdots cdots (364-N+1)}{364 times 364 times 364 cdots cdots 364 (N times)}$

But then how to proceed ?

help me

combinatorics number-theory

share|cite|improve this question

asked Mar 14 at 0:50

M. A. SARKARM. A. SARKAR

2,4291819

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a simplification but an approximation.
  $endgroup$
  – kimchi lover
  Mar 14 at 0:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kimchilover, How do I do this approximation ?
  $endgroup$
  – M. A. SARKAR
  Mar 14 at 0:52
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Use logarithms and the approximation $log(1-x) = -x -x^2/3cdots$ as explained in en.wikipedia.org/wiki/Birthday_problem
  $endgroup$
  – kimchi lover
  Mar 14 at 0:55
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Show that after simplification of $P=1-left( 364 times 363 cdots cdots frac{364-N+1}{364^N} right)$, we get the result $$P=1-e^{-frac{N(N-1)}{2 *364}}. $$

Answer:

We have,

$P=1-left( 364 times 363 cdots cdots frac{364-N+1}{364^N} right) \ Rightarrow P=1-frac{364 times 363 cdots cdots (364-N+1)}{364 times 364 times 364 cdots cdots 364 (N times)}$

But then how to proceed ?

help me

combinatorics number-theory

share|cite|improve this question

asked Mar 14 at 0:50

M. A. SARKARM. A. SARKAR

2,4291819

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a simplification but an approximation.
  $endgroup$
  – kimchi lover
  Mar 14 at 0:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kimchilover, How do I do this approximation ?
  $endgroup$
  – M. A. SARKAR
  Mar 14 at 0:52
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Use logarithms and the approximation $log(1-x) = -x -x^2/3cdots$ as explained in en.wikipedia.org/wiki/Birthday_problem
  $endgroup$
  – kimchi lover
  Mar 14 at 0:55
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Show that after simplification of $P=1-left( 364 times 363 cdots cdots frac{364-N+1}{364^N} right)$, we get the result $$P=1-e^{-frac{N(N-1)}{2 *364}}. $$

Answer:

We have,

$P=1-left( 364 times 363 cdots cdots frac{364-N+1}{364^N} right) \ Rightarrow P=1-frac{364 times 363 cdots cdots (364-N+1)}{364 times 364 times 364 cdots cdots 364 (N times)}$

But then how to proceed ?

help me

combinatorics number-theory

share|cite|improve this question

asked Mar 14 at 0:50

M. A. SARKARM. A. SARKAR

2,4291819

$endgroup$

share|cite|improve this question

asked Mar 14 at 0:50

M. A. SARKARM. A. SARKAR

2,4291819

share|cite|improve this question

asked Mar 14 at 0:50

M. A. SARKARM. A. SARKAR

2,4291819

asked Mar 14 at 0:50

M. A. SARKARM. A. SARKAR

2,4291819

asked Mar 14 at 0:50

M. A. SARKARM. A. SARKAR

2,4291819