From 6 different novels and 3 different dicti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Step $1$: Select $4$ novels from $6$ different novels in $^6C_4$ ways.$^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$ ways.Step $2$: Select $1$ dicti

Vedclass · Accepted Answer

at least $1000$

Vedclass · Answer

(C) Step $1$: Select $4$ novels from $6$ different novels in $^6C_4$ ways.$^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$ ways.Step $2$: Select $1$ dictionary from $3$ different dictionaries in $^3C_1$ ways.$^3C_1 = 3$ ways.Step $3$: Arrange the $4$ selected novels and $1$ dictionary in a row such that the dictionary is always in the middle.Since the dictionary is fixed in the middle,we only need to arrange the $4$ novels in the remaining $4$ positions.The number of ways to arrange $4$ novels is $4! = 24$.Step $4$: Total number of arrangements = $^6C_4 	imes ^3C_1 	imes 4! = 15 	imes 3 	imes 24 = 1080$.

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 arrangements is :

Similar Questions

The sum of all the $4$-digit numbers formed by taking all the digits from ${2, 3, 5, 7}$ without repetition is:

The number of integers between $1$ and $10^{10}$ (inclusive) that contain the digit $1$ is:

The number of all $3$-digit numbers $abc$ (in base $10$) for which $(a \times b \times c) + (a \times b) + (b \times c) + (c \times a) + a + b + c = 29$ is:

The number $(49^{2}-4)(49^{3}-49)$ is divisible by (in $!$)

The number of different $5$-digit numbers greater than $50000$ that can be formed using the digits $0, 1, 2, 3, 4, 5, 6, 7$,such that the sum of their first and last digits is not more than $8$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module