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(a) r = 4

(b) r = 3

(c) r = 7

(d) r = 8

Answer
Verified

We know that the a : b can also be expressed in the fractional form as $\dfrac{a}{b}$. So we get,

\[\dfrac{^{28}{{C}_{2r}}}{^{24}{{C}_{2r}}}=\dfrac{225}{11}\]

And the value of $^{n}{{C}_{r}}$ is given by $\dfrac{n!}{r!(n-r)!}$.

Now by expanding the expression, we get

\[\begin{align}

& \dfrac{^{28}{{C}_{2r}}}{^{24}{{C}_{2r}}}=\dfrac{225}{11} \\

& \dfrac{\dfrac{28!}{2r!(28-2r)!}}{\dfrac{24!}{2r!(24-2r)!}}=\dfrac{225}{11} \\

& \dfrac{28!\,\times \,(24-2r!)}{24!\,\times (28-2r!)}=\dfrac{225}{11} \\

\end{align}\]

By expanding the factorials, we get

\[\dfrac{28\times \,27\times 26\times 25\times 24!\times (24-2r)!}{24!\,\times (28-2r)\times (27-2r)\times (26-2r)\times (25-2r)\times (24-2r)!}=\dfrac{225}{11}\]

By cancelling out similar terms, we get

\[\dfrac{28\times 27\times 26\times 25}{(28-2r)(27-2r)(26-2r)(25-2r)}=\dfrac{225}{11}\]

After further simplifying, we get

\[(28-2r)(27-2r)(26-2r)(25-2r)=\dfrac{28\times 27\times 26\times 25\times 11}{225}\]

Simplifying the RHS of the above expression,

\[(28-2r)(27-2r)(26-2r)(25-2r)=14\times 13\times 12\times 11\]

Now, if we observe, in the above expression there are 4 consecutive terms on LHS and 4 consecutive terms on RHS so, we can directly compare both of the sides,

Hence by comparing, we get

$\begin{align}

& 28-2r=14 \\

& 2r=28-14 \\

& r=7 \\

\end{align}$

We can also compare $(27-2r)$ with 13 and so on and will get the same value of $r$ as above.