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. Then the number of such arrangements 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

Hint:
To solve this problem we use the method of combination. A combination is a group that can be made by taking some or all of the number of given things at a time. Since we have 6 different novels and 3 dictionaries. We have to choose 4 novels and one dictionary. so first we have to calculate the number of ways dictionaries can choose and the number of ways novels can be chosen. After that, we have to calculate the total number of arrangements.

Complete step by step solution:
Total number of novel s= 6
Number of dictionaries = 3
Now, we have to choose 4 novels from 6 novels
So the number of ways = $ ^6{C_4} $
We have to choose one dictionary from 3 dictionaries. The number of ways is $ ^3{C_1} $ .
Total number of ways in which 4 novels and one dictionary can be chosen = $ ^6{C_4} $ . $ ^3{C_1} $
Now, it is given that the arrangement dictionary is placed in the middle. So, we are left with four places In which we have to arrange the novels
\[\mathop \_\limits^ \downarrow ,\mathop \_\limits^ \downarrow ,\mathop \_\limits^D ,\mathop \_\limits^ \downarrow ,\mathop \_\limits^ \downarrow \]
Now the number of ways the four novels can be arranged in four places = $ 4! $ . Therefore total arrangements = total number of ways 4 novels and 1 dictionary is chosen × number of ways four novels can be arranged in four places
 $ { = ^6}{C_4}{ \times ^3}{C_1} \times 4! $
 $ = \dfrac{{6!}}{{4!(6 - 4)!}} \times \dfrac{{3!}}{{1!(3 - 1)!}} \times 4! $
 $ = \dfrac{{6 \times 5 \times 4!}}{{4! \times 2!}} \times \dfrac{{3!}}{{2!}} \times 4! $
 $ = \dfrac{{6 \times 5}}{2} \times \dfrac{{3 \times 2}}{2} \times 4 \times 3 \times 2 \times 1 $
15× 3× 4× 3× 2 = 1080
So, total number of arrangement = 1080

Hence, option (D) is correct.

Note:
 $ ^n{C_r} $ means the number of arrangements of n different things taken ‘r ‘ at a time.
Example – consider three letters a,b,c. The group of these three letters taken at a time is ab, bc, ca.
We concerted ac or ca in the same group as in a group we are concerned with the number of things contained.