Question

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangement is: -
(a) less than 500
(b) at least 500 but less than 750
(c) at least 750 but less than 1000
(d) at least 1000

Answer 1

Find the total number of ways to select 4 novels from 6 novels using the relation ${}^{6}{{C}_{4}}$. Now, calculate the total number of ways to select 1 dictionary from 3 dictionaries using the relation ${}^{3}{{C}_{1}}$. Now, arrange these at 5 places such that the dictionary always remains at the ${{3}^{rd}}$ place. Finally, multiply all the obtained ways to find the total number of arrangements possible.

Complete step-by-step solution
Here, we have been provided with 6 novels and 3 dictionaries and we have to select 4 novels and 1 dictionary from these.
Now, we know that the formula for selecting ‘r’ articles from a total of ‘n’ articles are given as: - ${}^{n}{{C}_{r}}$. This is the combination formula. So, we have,
Number of ways to select 4 novels from a total of 6 novels = ${}^{6}{{C}_{4}}$.
Number of ways to select 1 dictionary from a total of 3 dictionaries = ${}^{3}{{C}_{1}}$.
Now, the required number of novels and dictionaries are selected and now we have to arrange these in such a manner that the dictionary always remains in the middle.
Since we have 5 articles so we can arrange them at five places as shown below.

1	2	3	4	5

Since the dictionary should be in the middle, so it should always be at ${{3}^{rd}}$ place. Now, the four novels can be arranged at four different places, i.e. at 1, 2, 4, and 5.
Therefore, the number of ways to arrange these four novels at four different places = 4!
Now, the total number of arrangements possible will be the product of all the ways obtained. So, we have,
Total number of arrangements = ${}^{6}{{C}_{4}}\times {}^{3}{{C}_{1}}\times 4!$
Using the conversion, ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$, we get,
$\Rightarrow$ Total number of arrangements = $\dfrac{6!}{4!\times 2!}\times \dfrac{3!}{2!\times 1!}\times 4!$
$\Rightarrow$ Total number of arrangements = $\dfrac{6\times 5}{2}\times 3\times 4\times 3\times 2\times 1$
$\Rightarrow$ Total number of arrangements = 1080
Since, the total number of arrangements is 1080, i.e. at least 1000. Hence, option (d) is the correct answer.

Note: One may note that we have obtained the answer as 1080 arrangements and none of the options has this numerical value. This does not mean that the options are wrong because in option (d) it is said that the number of arrangements is at least 1000, which means the answer will be 1000 or more than that. Also, note that for selecting 4 novels and 1 dictionary, we do not have to apply the formula of permutation given as ${}^{n}{{P}_{r}}$.