Let's start with the given ratio:
[tex]\[ \frac{{}^{28}C_r}{{}^{26}C_{r-2}} = \frac{18}{5} \][/tex]
The combination formula is given by:
[tex]\[ {}^nC_r = \frac{n!}{r!(n-r)!} \][/tex]
Applying this to the given problem, we have:
[tex]\[ {}^{28}C_r = \frac{28!}{r!(28-r)!} \][/tex]
[tex]\[ {}^{26}C_{r-2} = \frac{26!}{(r-2)!(26-(r-2))!} = \frac{26!}{(r-2)!(28-r)!} \][/tex]
Now, substitute these into the ratio:
[tex]\[ \frac{\frac{28!}{r!(28-r)!}}{\frac{26!}{(r-2)!(28-r)!}} = \frac{18}{5} \][/tex]
Simplify the fraction by cancelling [tex]\( (28 - r)! \)[/tex] from the numerator and the denominator:
[tex]\[ \frac{28!}{r!(28-r)!} \cdot \frac{(r-2)!(28-r)!}{26!} = \frac{18}{5} \][/tex]
This becomes:
[tex]\[ \frac{28! \cdot (r-2)!}{r! \cdot 26!} = \frac{18}{5} \][/tex]
Factorize [tex]\( 28! \)[/tex] as [tex]\( 28 \times 27 \times 26! \)[/tex] and simplify:
[tex]\[ \frac{28 \times 27 \times 26! \cdot (r-2)!}{r \times (r-1) \times (r-2)! \cdot 26!} = \frac{18}{5} \][/tex]
Cancel [tex]\( 26! \)[/tex] and [tex]\( (r-2)! \)[/tex]:
[tex]\[ \frac{28 \times 27}{r(r-1)} = \frac{18}{5} \][/tex]
Cross-multiply to solve for [tex]\( r \)[/tex]:
[tex]\[ 28 \times 27 \times 5 = 18 \times r(r-1) \][/tex]
This simplifies to:
[tex]\[ 3780 = 18r^2 - 18r \][/tex]
Divide through by 18:
[tex]\[ 210 = r^2 - r \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ r^2 - r - 210 = 0 \][/tex]
Solve the quadratic equation using the quadratic formula [tex]\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -210 \)[/tex]:
[tex]\[ r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-210)}}{2 \cdot 1} \][/tex]
[tex]\[ r = \frac{1 \pm \sqrt{1 + 840}}{2} \][/tex]
[tex]\[ r = \frac{1 \pm \sqrt{841}}{2} \][/tex]
[tex]\[ r = \frac{1 \pm 29}{2} \][/tex]
This gives us two possible solutions:
[tex]\[ r = \frac{1 + 29}{2} = 15 \][/tex]
[tex]\[ r = \frac{1 - 29}{2} = -14 \][/tex]
Since [tex]\( r \)[/tex] must be a non-negative integer (as it represents a number of combinations), we discard [tex]\( r = -14 \)[/tex].
Therefore, the value of [tex]\( r \)[/tex] is:
[tex]\[ r = 15 \][/tex]
