Questions & Answers

Find the value of r, if we are given that \[^{28}{{C}_{2r}}{{:}^{24}}{{C}_{2r}}=225 :11\].
(a) r = 4
(b) r = 3
(c) r = 7
(d) r = 8

Answer Verified Verified
Hint: To solve this question we need to first know that a: b can also be expressed in the fractional form as $\dfrac{a}{b}$ and the value of $^{n}{{C}_{r}}$ is given by $\dfrac{n!}{r!(n-r)!}$. Now after that, we will put the required values in the given expression and solve them to find the value of r.

Complete step-by-step solution:
We know that the a : b can also be expressed in the fractional form as $\dfrac{a}{b}$. So we get,
And the value of $^{n}{{C}_{r}}$ is given by $\dfrac{n!}{r!(n-r)!}$.
Now by expanding the expression, we get
  & \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} \\
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
  & 28-2r=14 \\
 & 2r=28-14 \\
 & r=7 \\
We can also compare $(27-2r)$ with 13 and so on and will get the same value of $r$ as above.
Hence, value of $r$ we got is 7 which matches the option (c), hence (c) is the correct answer.

Note: You can also solve this question by putting each of the value of $r$ given in options and if LHS comes out to be equal to RHS it will be the correct answer but, this method involves a lot of calculations, and can be a little hectic. In the above solution, we have manipulated the RHS of the equation in such a way that we can directly compare it with LHS which makes the solving a lot easier so, whenever there is a polynomial expression (like in LHS in this question) which cannot be solved easily try to express RHS in such a way that it can be directly compared to LHS of the expression.