transformation of an equation (Binomial coefficient) [closed]

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Good afternoon



I don't understand this transformation. Can you please help me?



Greetings



Ava



$beginalignn choose k&=fracn1cdotfracn-12cdotsfracn-(k-1)k\&=fracncdot (n-1)cdots(n-k+1)k!endalign$










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closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, Theoretical Economist, Adrian Keister Sep 9 at 0:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Theoretical Economist, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.












  • the above part is just a multiplication $n-(k-1)=n-1cdot(k-1)=n-k+1$... The below part is just the definition of $k!$
    – koleygr
    Sep 8 at 12:19











  • Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
    – N. F. Taussig
    Sep 8 at 13:29










  • @N.F.Taussig Is it possible to remove the central = between lines by Latex?
    – Narasimham
    Sep 8 at 23:03










  • @Narasimham Somebody left an extra equals sign after the aligned equations, producing the central equals sign you noticed.
    – N. F. Taussig
    Sep 8 at 23:16














up vote
0
down vote

favorite












Good afternoon



I don't understand this transformation. Can you please help me?



Greetings



Ava



$beginalignn choose k&=fracn1cdotfracn-12cdotsfracn-(k-1)k\&=fracncdot (n-1)cdots(n-k+1)k!endalign$










share|cite|improve this question















closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, Theoretical Economist, Adrian Keister Sep 9 at 0:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Theoretical Economist, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.












  • the above part is just a multiplication $n-(k-1)=n-1cdot(k-1)=n-k+1$... The below part is just the definition of $k!$
    – koleygr
    Sep 8 at 12:19











  • Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
    – N. F. Taussig
    Sep 8 at 13:29










  • @N.F.Taussig Is it possible to remove the central = between lines by Latex?
    – Narasimham
    Sep 8 at 23:03










  • @Narasimham Somebody left an extra equals sign after the aligned equations, producing the central equals sign you noticed.
    – N. F. Taussig
    Sep 8 at 23:16












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Good afternoon



I don't understand this transformation. Can you please help me?



Greetings



Ava



$beginalignn choose k&=fracn1cdotfracn-12cdotsfracn-(k-1)k\&=fracncdot (n-1)cdots(n-k+1)k!endalign$










share|cite|improve this question















Good afternoon



I don't understand this transformation. Can you please help me?



Greetings



Ava



$beginalignn choose k&=fracn1cdotfracn-12cdotsfracn-(k-1)k\&=fracncdot (n-1)cdots(n-k+1)k!endalign$







binomial-coefficients transformation






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edited Sep 8 at 23:15









N. F. Taussig

39.7k93153




39.7k93153










asked Sep 8 at 11:46







user591158











closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, Theoretical Economist, Adrian Keister Sep 9 at 0:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Theoretical Economist, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Jyrki Lahtonen, José Carlos Santos, Theoretical Economist, Adrian Keister Sep 9 at 0:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Theoretical Economist, Adrian Keister
If this question can be reworded to fit the rules in the help center, please edit the question.











  • the above part is just a multiplication $n-(k-1)=n-1cdot(k-1)=n-k+1$... The below part is just the definition of $k!$
    – koleygr
    Sep 8 at 12:19











  • Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
    – N. F. Taussig
    Sep 8 at 13:29










  • @N.F.Taussig Is it possible to remove the central = between lines by Latex?
    – Narasimham
    Sep 8 at 23:03










  • @Narasimham Somebody left an extra equals sign after the aligned equations, producing the central equals sign you noticed.
    – N. F. Taussig
    Sep 8 at 23:16
















  • the above part is just a multiplication $n-(k-1)=n-1cdot(k-1)=n-k+1$... The below part is just the definition of $k!$
    – koleygr
    Sep 8 at 12:19











  • Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
    – N. F. Taussig
    Sep 8 at 13:29










  • @N.F.Taussig Is it possible to remove the central = between lines by Latex?
    – Narasimham
    Sep 8 at 23:03










  • @Narasimham Somebody left an extra equals sign after the aligned equations, producing the central equals sign you noticed.
    – N. F. Taussig
    Sep 8 at 23:16















the above part is just a multiplication $n-(k-1)=n-1cdot(k-1)=n-k+1$... The below part is just the definition of $k!$
– koleygr
Sep 8 at 12:19





the above part is just a multiplication $n-(k-1)=n-1cdot(k-1)=n-k+1$... The below part is just the definition of $k!$
– koleygr
Sep 8 at 12:19













Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
– N. F. Taussig
Sep 8 at 13:29




Welcome to MathSE. Please read this tutorial, which explains how to typeset mathematics on this site using MathJax.
– N. F. Taussig
Sep 8 at 13:29












@N.F.Taussig Is it possible to remove the central = between lines by Latex?
– Narasimham
Sep 8 at 23:03




@N.F.Taussig Is it possible to remove the central = between lines by Latex?
– Narasimham
Sep 8 at 23:03












@Narasimham Somebody left an extra equals sign after the aligned equations, producing the central equals sign you noticed.
– N. F. Taussig
Sep 8 at 23:16




@Narasimham Somebody left an extra equals sign after the aligned equations, producing the central equals sign you noticed.
– N. F. Taussig
Sep 8 at 23:16










1 Answer
1






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up vote
1
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By definition $$binomnk=fracn!k!(n-k)!=fracn(n-1)ldots(n-k+1)(n-k)!k!(n-k)!$$



Cancel out the$ (n-k)!$ and you're done.






share|cite|improve this answer




















  • Good evening But I don't understand how to get from n factorial to this equation in the numerator. greetings ava
    – user591158
    Sep 8 at 16:34










  • Good evening @Ava .. Well imagine $k=3$ and $n=5$.. couldn't you write $5!$ as $5.4.3.2!$ where $2!=(n-k)!$?
    – Ahmad Bazzi
    Sep 8 at 17:42











  • But n factorial should be n(n-1)(n-2).... But why n(n-1)(n-2)...(n-k+1)(n-k)factorial
    – user591158
    Sep 8 at 18:05











  • Mathematically $n(n-1)(n-2)....(2)(1) = n(n-1)(n-2)...(n-k+1)(n-k)!$
    – Ahmad Bazzi
    Sep 8 at 19:42
















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













By definition $$binomnk=fracn!k!(n-k)!=fracn(n-1)ldots(n-k+1)(n-k)!k!(n-k)!$$



Cancel out the$ (n-k)!$ and you're done.






share|cite|improve this answer




















  • Good evening But I don't understand how to get from n factorial to this equation in the numerator. greetings ava
    – user591158
    Sep 8 at 16:34










  • Good evening @Ava .. Well imagine $k=3$ and $n=5$.. couldn't you write $5!$ as $5.4.3.2!$ where $2!=(n-k)!$?
    – Ahmad Bazzi
    Sep 8 at 17:42











  • But n factorial should be n(n-1)(n-2).... But why n(n-1)(n-2)...(n-k+1)(n-k)factorial
    – user591158
    Sep 8 at 18:05











  • Mathematically $n(n-1)(n-2)....(2)(1) = n(n-1)(n-2)...(n-k+1)(n-k)!$
    – Ahmad Bazzi
    Sep 8 at 19:42














up vote
1
down vote













By definition $$binomnk=fracn!k!(n-k)!=fracn(n-1)ldots(n-k+1)(n-k)!k!(n-k)!$$



Cancel out the$ (n-k)!$ and you're done.






share|cite|improve this answer




















  • Good evening But I don't understand how to get from n factorial to this equation in the numerator. greetings ava
    – user591158
    Sep 8 at 16:34










  • Good evening @Ava .. Well imagine $k=3$ and $n=5$.. couldn't you write $5!$ as $5.4.3.2!$ where $2!=(n-k)!$?
    – Ahmad Bazzi
    Sep 8 at 17:42











  • But n factorial should be n(n-1)(n-2).... But why n(n-1)(n-2)...(n-k+1)(n-k)factorial
    – user591158
    Sep 8 at 18:05











  • Mathematically $n(n-1)(n-2)....(2)(1) = n(n-1)(n-2)...(n-k+1)(n-k)!$
    – Ahmad Bazzi
    Sep 8 at 19:42












up vote
1
down vote










up vote
1
down vote









By definition $$binomnk=fracn!k!(n-k)!=fracn(n-1)ldots(n-k+1)(n-k)!k!(n-k)!$$



Cancel out the$ (n-k)!$ and you're done.






share|cite|improve this answer












By definition $$binomnk=fracn!k!(n-k)!=fracn(n-1)ldots(n-k+1)(n-k)!k!(n-k)!$$



Cancel out the$ (n-k)!$ and you're done.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 8 at 13:23









Ahmad Bazzi

6,4061624




6,4061624











  • Good evening But I don't understand how to get from n factorial to this equation in the numerator. greetings ava
    – user591158
    Sep 8 at 16:34










  • Good evening @Ava .. Well imagine $k=3$ and $n=5$.. couldn't you write $5!$ as $5.4.3.2!$ where $2!=(n-k)!$?
    – Ahmad Bazzi
    Sep 8 at 17:42











  • But n factorial should be n(n-1)(n-2).... But why n(n-1)(n-2)...(n-k+1)(n-k)factorial
    – user591158
    Sep 8 at 18:05











  • Mathematically $n(n-1)(n-2)....(2)(1) = n(n-1)(n-2)...(n-k+1)(n-k)!$
    – Ahmad Bazzi
    Sep 8 at 19:42
















  • Good evening But I don't understand how to get from n factorial to this equation in the numerator. greetings ava
    – user591158
    Sep 8 at 16:34










  • Good evening @Ava .. Well imagine $k=3$ and $n=5$.. couldn't you write $5!$ as $5.4.3.2!$ where $2!=(n-k)!$?
    – Ahmad Bazzi
    Sep 8 at 17:42











  • But n factorial should be n(n-1)(n-2).... But why n(n-1)(n-2)...(n-k+1)(n-k)factorial
    – user591158
    Sep 8 at 18:05











  • Mathematically $n(n-1)(n-2)....(2)(1) = n(n-1)(n-2)...(n-k+1)(n-k)!$
    – Ahmad Bazzi
    Sep 8 at 19:42















Good evening But I don't understand how to get from n factorial to this equation in the numerator. greetings ava
– user591158
Sep 8 at 16:34




Good evening But I don't understand how to get from n factorial to this equation in the numerator. greetings ava
– user591158
Sep 8 at 16:34












Good evening @Ava .. Well imagine $k=3$ and $n=5$.. couldn't you write $5!$ as $5.4.3.2!$ where $2!=(n-k)!$?
– Ahmad Bazzi
Sep 8 at 17:42





Good evening @Ava .. Well imagine $k=3$ and $n=5$.. couldn't you write $5!$ as $5.4.3.2!$ where $2!=(n-k)!$?
– Ahmad Bazzi
Sep 8 at 17:42













But n factorial should be n(n-1)(n-2).... But why n(n-1)(n-2)...(n-k+1)(n-k)factorial
– user591158
Sep 8 at 18:05





But n factorial should be n(n-1)(n-2).... But why n(n-1)(n-2)...(n-k+1)(n-k)factorial
– user591158
Sep 8 at 18:05













Mathematically $n(n-1)(n-2)....(2)(1) = n(n-1)(n-2)...(n-k+1)(n-k)!$
– Ahmad Bazzi
Sep 8 at 19:42




Mathematically $n(n-1)(n-2)....(2)(1) = n(n-1)(n-2)...(n-k+1)(n-k)!$
– Ahmad Bazzi
Sep 8 at 19:42


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