Show that the number of sequences of length 40 are there using the alphabet {a,b,c,d}












1














Show that the number of sequences of length 40 are there using the alphabet {a,b,c,d} such that number of a's in the sequence is divisible by 3 is $frac{4^{40}+2Re((3+w)^{40})}{3}$



sequence with 3 a's=$40C3times3^{(40-3)}$



sequence with 6 a's=$40C6times 3^{(40-6)}$



sequence with 9 a's=$40C9times3^{(40-9)}$



.



.



sequence with 36 a's=$40C36times3^{(40-36)}$
sequence with 39 a's=$40C39times3^{(40-39)}$



I'm stuck as to where the Re and $w$ come into play






combinatorics







share|cite|improve this question


















  • 1




    The $w$ might be $omega$, a unit cube root in $mathbb{C}$..
    – Henno Brandsma
    Dec 2 at 10:29






  • 1




    Use the same technique as here. Instead of $(1+x)^{40}$ with $x=1,omega,omega^2$, you will be needing $(3+x)^{40}$ with $x=1,omega,omega^2$. Do you see why?
    – Jyrki Lahtonen
    Dec 2 at 10:34












  • Yes after the solution down there , I now fully understand
    – Tariro Manyika
    Dec 2 at 21:11










  • That's great. TariroManyika !
    – Jyrki Lahtonen
    Dec 5 at 19:47
















1














Show that the number of sequences of length 40 are there using the alphabet {a,b,c,d} such that number of a's in the sequence is divisible by 3 is $frac{4^{40}+2Re((3+w)^{40})}{3}$



sequence with 3 a's=$40C3times3^{(40-3)}$



sequence with 6 a's=$40C6times 3^{(40-6)}$



sequence with 9 a's=$40C9times3^{(40-9)}$



.



.



sequence with 36 a's=$40C36times3^{(40-36)}$
sequence with 39 a's=$40C39times3^{(40-39)}$



I'm stuck as to where the Re and $w$ come into play






combinatorics







share|cite|improve this question


















  • 1




    The $w$ might be $omega$, a unit cube root in $mathbb{C}$..
    – Henno Brandsma
    Dec 2 at 10:29






  • 1




    Use the same technique as here. Instead of $(1+x)^{40}$ with $x=1,omega,omega^2$, you will be needing $(3+x)^{40}$ with $x=1,omega,omega^2$. Do you see why?
    – Jyrki Lahtonen
    Dec 2 at 10:34












  • Yes after the solution down there , I now fully understand
    – Tariro Manyika
    Dec 2 at 21:11










  • That's great. TariroManyika !
    – Jyrki Lahtonen
    Dec 5 at 19:47














1












1








1


0





Show that the number of sequences of length 40 are there using the alphabet {a,b,c,d} such that number of a's in the sequence is divisible by 3 is $frac{4^{40}+2Re((3+w)^{40})}{3}$



sequence with 3 a's=$40C3times3^{(40-3)}$



sequence with 6 a's=$40C6times 3^{(40-6)}$



sequence with 9 a's=$40C9times3^{(40-9)}$



.



.



sequence with 36 a's=$40C36times3^{(40-36)}$
sequence with 39 a's=$40C39times3^{(40-39)}$



I'm stuck as to where the Re and $w$ come into play






combinatorics







share|cite|improve this question













Show that the number of sequences of length 40 are there using the alphabet {a,b,c,d} such that number of a's in the sequence is divisible by 3 is $frac{4^{40}+2Re((3+w)^{40})}{3}$



sequence with 3 a's=$40C3times3^{(40-3)}$



sequence with 6 a's=$40C6times 3^{(40-6)}$



sequence with 9 a's=$40C9times3^{(40-9)}$



.



.



sequence with 36 a's=$40C36times3^{(40-36)}$
sequence with 39 a's=$40C39times3^{(40-39)}$



I'm stuck as to where the Re and $w$ come into play






combinatorics




combinatorics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 at 10:20









Tariro Manyika

62




62








  • 1




    The $w$ might be $omega$, a unit cube root in $mathbb{C}$..
    – Henno Brandsma
    Dec 2 at 10:29






  • 1




    Use the same technique as here. Instead of $(1+x)^{40}$ with $x=1,omega,omega^2$, you will be needing $(3+x)^{40}$ with $x=1,omega,omega^2$. Do you see why?
    – Jyrki Lahtonen
    Dec 2 at 10:34












  • Yes after the solution down there , I now fully understand
    – Tariro Manyika
    Dec 2 at 21:11










  • That's great. TariroManyika !
    – Jyrki Lahtonen
    Dec 5 at 19:47














  • 1




    The $w$ might be $omega$, a unit cube root in $mathbb{C}$..
    – Henno Brandsma
    Dec 2 at 10:29






  • 1




    Use the same technique as here. Instead of $(1+x)^{40}$ with $x=1,omega,omega^2$, you will be needing $(3+x)^{40}$ with $x=1,omega,omega^2$. Do you see why?
    – Jyrki Lahtonen
    Dec 2 at 10:34












  • Yes after the solution down there , I now fully understand
    – Tariro Manyika
    Dec 2 at 21:11










  • That's great. TariroManyika !
    – Jyrki Lahtonen
    Dec 5 at 19:47








1




1




The $w$ might be $omega$, a unit cube root in $mathbb{C}$..
– Henno Brandsma
Dec 2 at 10:29




The $w$ might be $omega$, a unit cube root in $mathbb{C}$..
– Henno Brandsma
Dec 2 at 10:29




1




1




Use the same technique as here. Instead of $(1+x)^{40}$ with $x=1,omega,omega^2$, you will be needing $(3+x)^{40}$ with $x=1,omega,omega^2$. Do you see why?
– Jyrki Lahtonen
Dec 2 at 10:34






Use the same technique as here. Instead of $(1+x)^{40}$ with $x=1,omega,omega^2$, you will be needing $(3+x)^{40}$ with $x=1,omega,omega^2$. Do you see why?
– Jyrki Lahtonen
Dec 2 at 10:34














Yes after the solution down there , I now fully understand
– Tariro Manyika
Dec 2 at 21:11




Yes after the solution down there , I now fully understand
– Tariro Manyika
Dec 2 at 21:11












That's great. TariroManyika !
– Jyrki Lahtonen
Dec 5 at 19:47




That's great. TariroManyika !
– Jyrki Lahtonen
Dec 5 at 19:47










1 Answer
1






active

oldest

votes


















2














Consider the polynomial $(a+b+c+d)^{40}$. The coefficient of a term of the form $a^{k_1}b^{k_2}c^{k_3}d^{k_4}$ counts the number of sequences of length 40 composed of ${k_1}$ $a$'s, ${k_2}$ $b$'s, ${k_3}$ $c$'s, ${k_2}$ $d$'s. Therefore we are required to find the sum of coefficients of the terms of $(a+b+c+d)^{40}$ with the exponent of a divisible by the $3$.



To do this, we set $b=c=d=1$ in the polynomial as there is no restriction on the number of $b$'s, $c$'s or $d$'s. Thus our answer is the sum of coefficients of the terms with exponent divisible by $3$ of the polynomial $(a+3)^{40}$.



Let $f(a)=(a+3)^{40}=sum_{i=0}^{40}{l_i}a^{i}$. Therefore we have to get hold of $z=l_0+l_3+l_6+...+l_{39}$. Notice $f(1)= l_0+l_1+l_2+...+l_{40}$, $f(omega)=l_0+l_1omega+l_2omega^{2}+l_3+l_4omega+...+l_{40}omega$. $f(omega^2)=l_0+l_1omega^{2}+l_2omega+l_3+l_4omega^{2}+...+l_{40}omega^{2}$.
Therefore $f(1)+f(omega)+f(omega^{2})=3(l_0+l_3+l_6+...+l_{39})=3z$



Hence, $$z=frac{4^{40}+(3+omega)^{40}+(3+omega^{2})^{40}}{3}
=frac{4^{40}+(3+omega)^{40}+(3+bar omega)^{40}}{3}
=frac{4^{40}+2Re((3+omega)^{40})}{3}$$
Q.E.D.






share|cite|improve this answer























  • Thank you so much , which text book can I read that can help me answer such questions in future ?
    – Tariro Manyika
    Dec 2 at 21:11










  • Sorry after hence, If w is a complex number, (3+w)^2 not equal to (3+conjugate w)
    – Tariro Manyika
    Dec 2 at 23:46










  • A Path to Combinatorics for Undergraduates by Titu Andreescu and Zuming Feng is good for general reference. I would recommend Generatingfunctionology by Herbert Wilf for exposure to a wide spectrum of problems solved using generating functions.
    – Anubhab Ghosal
    Dec 3 at 3:10










  • As $omega^{3}=1$, $bar omega = frac{1}{omega} = omega^{2}$.
    – Anubhab Ghosal
    Dec 3 at 3:14













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Consider the polynomial $(a+b+c+d)^{40}$. The coefficient of a term of the form $a^{k_1}b^{k_2}c^{k_3}d^{k_4}$ counts the number of sequences of length 40 composed of ${k_1}$ $a$'s, ${k_2}$ $b$'s, ${k_3}$ $c$'s, ${k_2}$ $d$'s. Therefore we are required to find the sum of coefficients of the terms of $(a+b+c+d)^{40}$ with the exponent of a divisible by the $3$.



To do this, we set $b=c=d=1$ in the polynomial as there is no restriction on the number of $b$'s, $c$'s or $d$'s. Thus our answer is the sum of coefficients of the terms with exponent divisible by $3$ of the polynomial $(a+3)^{40}$.



Let $f(a)=(a+3)^{40}=sum_{i=0}^{40}{l_i}a^{i}$. Therefore we have to get hold of $z=l_0+l_3+l_6+...+l_{39}$. Notice $f(1)= l_0+l_1+l_2+...+l_{40}$, $f(omega)=l_0+l_1omega+l_2omega^{2}+l_3+l_4omega+...+l_{40}omega$. $f(omega^2)=l_0+l_1omega^{2}+l_2omega+l_3+l_4omega^{2}+...+l_{40}omega^{2}$.
Therefore $f(1)+f(omega)+f(omega^{2})=3(l_0+l_3+l_6+...+l_{39})=3z$



Hence, $$z=frac{4^{40}+(3+omega)^{40}+(3+omega^{2})^{40}}{3}
=frac{4^{40}+(3+omega)^{40}+(3+bar omega)^{40}}{3}
=frac{4^{40}+2Re((3+omega)^{40})}{3}$$
Q.E.D.






share|cite|improve this answer























  • Thank you so much , which text book can I read that can help me answer such questions in future ?
    – Tariro Manyika
    Dec 2 at 21:11










  • Sorry after hence, If w is a complex number, (3+w)^2 not equal to (3+conjugate w)
    – Tariro Manyika
    Dec 2 at 23:46










  • A Path to Combinatorics for Undergraduates by Titu Andreescu and Zuming Feng is good for general reference. I would recommend Generatingfunctionology by Herbert Wilf for exposure to a wide spectrum of problems solved using generating functions.
    – Anubhab Ghosal
    Dec 3 at 3:10










  • As $omega^{3}=1$, $bar omega = frac{1}{omega} = omega^{2}$.
    – Anubhab Ghosal
    Dec 3 at 3:14


















2














Consider the polynomial $(a+b+c+d)^{40}$. The coefficient of a term of the form $a^{k_1}b^{k_2}c^{k_3}d^{k_4}$ counts the number of sequences of length 40 composed of ${k_1}$ $a$'s, ${k_2}$ $b$'s, ${k_3}$ $c$'s, ${k_2}$ $d$'s. Therefore we are required to find the sum of coefficients of the terms of $(a+b+c+d)^{40}$ with the exponent of a divisible by the $3$.



To do this, we set $b=c=d=1$ in the polynomial as there is no restriction on the number of $b$'s, $c$'s or $d$'s. Thus our answer is the sum of coefficients of the terms with exponent divisible by $3$ of the polynomial $(a+3)^{40}$.



Let $f(a)=(a+3)^{40}=sum_{i=0}^{40}{l_i}a^{i}$. Therefore we have to get hold of $z=l_0+l_3+l_6+...+l_{39}$. Notice $f(1)= l_0+l_1+l_2+...+l_{40}$, $f(omega)=l_0+l_1omega+l_2omega^{2}+l_3+l_4omega+...+l_{40}omega$. $f(omega^2)=l_0+l_1omega^{2}+l_2omega+l_3+l_4omega^{2}+...+l_{40}omega^{2}$.
Therefore $f(1)+f(omega)+f(omega^{2})=3(l_0+l_3+l_6+...+l_{39})=3z$



Hence, $$z=frac{4^{40}+(3+omega)^{40}+(3+omega^{2})^{40}}{3}
=frac{4^{40}+(3+omega)^{40}+(3+bar omega)^{40}}{3}
=frac{4^{40}+2Re((3+omega)^{40})}{3}$$
Q.E.D.






share|cite|improve this answer























  • Thank you so much , which text book can I read that can help me answer such questions in future ?
    – Tariro Manyika
    Dec 2 at 21:11










  • Sorry after hence, If w is a complex number, (3+w)^2 not equal to (3+conjugate w)
    – Tariro Manyika
    Dec 2 at 23:46










  • A Path to Combinatorics for Undergraduates by Titu Andreescu and Zuming Feng is good for general reference. I would recommend Generatingfunctionology by Herbert Wilf for exposure to a wide spectrum of problems solved using generating functions.
    – Anubhab Ghosal
    Dec 3 at 3:10










  • As $omega^{3}=1$, $bar omega = frac{1}{omega} = omega^{2}$.
    – Anubhab Ghosal
    Dec 3 at 3:14
















2












2








2






Consider the polynomial $(a+b+c+d)^{40}$. The coefficient of a term of the form $a^{k_1}b^{k_2}c^{k_3}d^{k_4}$ counts the number of sequences of length 40 composed of ${k_1}$ $a$'s, ${k_2}$ $b$'s, ${k_3}$ $c$'s, ${k_2}$ $d$'s. Therefore we are required to find the sum of coefficients of the terms of $(a+b+c+d)^{40}$ with the exponent of a divisible by the $3$.



To do this, we set $b=c=d=1$ in the polynomial as there is no restriction on the number of $b$'s, $c$'s or $d$'s. Thus our answer is the sum of coefficients of the terms with exponent divisible by $3$ of the polynomial $(a+3)^{40}$.



Let $f(a)=(a+3)^{40}=sum_{i=0}^{40}{l_i}a^{i}$. Therefore we have to get hold of $z=l_0+l_3+l_6+...+l_{39}$. Notice $f(1)= l_0+l_1+l_2+...+l_{40}$, $f(omega)=l_0+l_1omega+l_2omega^{2}+l_3+l_4omega+...+l_{40}omega$. $f(omega^2)=l_0+l_1omega^{2}+l_2omega+l_3+l_4omega^{2}+...+l_{40}omega^{2}$.
Therefore $f(1)+f(omega)+f(omega^{2})=3(l_0+l_3+l_6+...+l_{39})=3z$



Hence, $$z=frac{4^{40}+(3+omega)^{40}+(3+omega^{2})^{40}}{3}
=frac{4^{40}+(3+omega)^{40}+(3+bar omega)^{40}}{3}
=frac{4^{40}+2Re((3+omega)^{40})}{3}$$
Q.E.D.






share|cite|improve this answer














Consider the polynomial $(a+b+c+d)^{40}$. The coefficient of a term of the form $a^{k_1}b^{k_2}c^{k_3}d^{k_4}$ counts the number of sequences of length 40 composed of ${k_1}$ $a$'s, ${k_2}$ $b$'s, ${k_3}$ $c$'s, ${k_2}$ $d$'s. Therefore we are required to find the sum of coefficients of the terms of $(a+b+c+d)^{40}$ with the exponent of a divisible by the $3$.



To do this, we set $b=c=d=1$ in the polynomial as there is no restriction on the number of $b$'s, $c$'s or $d$'s. Thus our answer is the sum of coefficients of the terms with exponent divisible by $3$ of the polynomial $(a+3)^{40}$.



Let $f(a)=(a+3)^{40}=sum_{i=0}^{40}{l_i}a^{i}$. Therefore we have to get hold of $z=l_0+l_3+l_6+...+l_{39}$. Notice $f(1)= l_0+l_1+l_2+...+l_{40}$, $f(omega)=l_0+l_1omega+l_2omega^{2}+l_3+l_4omega+...+l_{40}omega$. $f(omega^2)=l_0+l_1omega^{2}+l_2omega+l_3+l_4omega^{2}+...+l_{40}omega^{2}$.
Therefore $f(1)+f(omega)+f(omega^{2})=3(l_0+l_3+l_6+...+l_{39})=3z$



Hence, $$z=frac{4^{40}+(3+omega)^{40}+(3+omega^{2})^{40}}{3}
=frac{4^{40}+(3+omega)^{40}+(3+bar omega)^{40}}{3}
=frac{4^{40}+2Re((3+omega)^{40})}{3}$$
Q.E.D.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 at 3:02

























answered Dec 2 at 13:47









Anubhab Ghosal

74412




74412












  • Thank you so much , which text book can I read that can help me answer such questions in future ?
    – Tariro Manyika
    Dec 2 at 21:11










  • Sorry after hence, If w is a complex number, (3+w)^2 not equal to (3+conjugate w)
    – Tariro Manyika
    Dec 2 at 23:46










  • A Path to Combinatorics for Undergraduates by Titu Andreescu and Zuming Feng is good for general reference. I would recommend Generatingfunctionology by Herbert Wilf for exposure to a wide spectrum of problems solved using generating functions.
    – Anubhab Ghosal
    Dec 3 at 3:10










  • As $omega^{3}=1$, $bar omega = frac{1}{omega} = omega^{2}$.
    – Anubhab Ghosal
    Dec 3 at 3:14




















  • Thank you so much , which text book can I read that can help me answer such questions in future ?
    – Tariro Manyika
    Dec 2 at 21:11










  • Sorry after hence, If w is a complex number, (3+w)^2 not equal to (3+conjugate w)
    – Tariro Manyika
    Dec 2 at 23:46










  • A Path to Combinatorics for Undergraduates by Titu Andreescu and Zuming Feng is good for general reference. I would recommend Generatingfunctionology by Herbert Wilf for exposure to a wide spectrum of problems solved using generating functions.
    – Anubhab Ghosal
    Dec 3 at 3:10










  • As $omega^{3}=1$, $bar omega = frac{1}{omega} = omega^{2}$.
    – Anubhab Ghosal
    Dec 3 at 3:14


















Thank you so much , which text book can I read that can help me answer such questions in future ?
– Tariro Manyika
Dec 2 at 21:11




Thank you so much , which text book can I read that can help me answer such questions in future ?
– Tariro Manyika
Dec 2 at 21:11












Sorry after hence, If w is a complex number, (3+w)^2 not equal to (3+conjugate w)
– Tariro Manyika
Dec 2 at 23:46




Sorry after hence, If w is a complex number, (3+w)^2 not equal to (3+conjugate w)
– Tariro Manyika
Dec 2 at 23:46












A Path to Combinatorics for Undergraduates by Titu Andreescu and Zuming Feng is good for general reference. I would recommend Generatingfunctionology by Herbert Wilf for exposure to a wide spectrum of problems solved using generating functions.
– Anubhab Ghosal
Dec 3 at 3:10




A Path to Combinatorics for Undergraduates by Titu Andreescu and Zuming Feng is good for general reference. I would recommend Generatingfunctionology by Herbert Wilf for exposure to a wide spectrum of problems solved using generating functions.
– Anubhab Ghosal
Dec 3 at 3:10












As $omega^{3}=1$, $bar omega = frac{1}{omega} = omega^{2}$.
– Anubhab Ghosal
Dec 3 at 3:14






As $omega^{3}=1$, $bar omega = frac{1}{omega} = omega^{2}$.
– Anubhab Ghosal
Dec 3 at 3:14




















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Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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