pascal terms

[red and white kop!]

ok so i'm using the notation (n r) for a pascal term

state numbers a, b and c such that
a*(8 5)=b*(8 6)=c*(7 5)

i thought this should be easy: i made three separate equation and developed the pascal terms into factorials like (8 5) = 8!/5!*3! = 8x7x6/2x3 but i just end up with stuff i don't know how to develop. how would you do this? i think there's just something simple im missing

 

[Deleted member 4993]

ok so i'm using the notation (n r) for a pascal term

state numbers a, b and c such that
a*(8 5)=b*(8 6)=c*(7 5)

i thought this should be easy: i made three separate equation and developed the pascal terms into factorials like (8 5) = 8!/(5!*3!) = 8x7x6/(2x3) = 56 but i just end up with stuff i don't know how to develop. how would you do this? i think there's just something simple im missing

56 a = 28b = 21c

You have two equations and three unknowns. So you can express those as ratios:

a : b : c = 1 : 2 : (8/3)

 

[soroban]

Hello, red and white kop!

There is an infinite number of solutions.

\(\displaystyle \text{Find numbers }a,\,b,\,c \text{ such that: }\;a{8\choose5} \:=\:b{8\choose6} \:=\:c{7\choose5}\)

\(\displaystyle \text{We are told: }\;a\,\cdot\frac{8!}{5!\,3!} \;=\;b\,\cdot\frac{8!}{6!\,2!} \;=\;c\,\cdot\frac{7!}{5!\,2!} \qquad\Rightarrow\qquad a\,\cdot\frac{8\cdot7\cdot6}{3\cdot2\cdot1} \;=\;b\,\cdot\frac{8\cdot7}{2\cdot1} \;=\;c\,\cdot\frac{7\cdot6}{2\cdot1}\)

. . . . . . . . . . \(\displaystyle 56a \;=\;28b \;=\;21c \qquad\Rightarrow\qquad 8a \;=\;4b \;=\;3c\)


\(\displaystyle \text{Then we have: }\;\begin{Bmatrix}8a\:=\:3c &\Rightarrow& a \:=\:\frac{3}{8}c \\ \\[-3mm]4b \:=\:3c &\Rightarrow& b \:=\:\frac{3}{4}c \\ \\[-4mm]& & c \:=\;\;\;c \end{Bmatrix}\)


\(\displaystyle \text{On the right, replace }c\text{ with }8t.\)

. . \(\displaystyle \text{Therefore: }\;\begin{Bmatrix}a &=&3t \\ b &=& 6t \\ c &=& 8t \end{Bmatrix}\;\;\text{ for any real number, } t\)