Out of 6 distinct novels and 3 distinct dicti | vedclass

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

Question

(B) The number of ways to select $4$ novels from $6$ is $\binom{6}{4} = 15$.The number of ways to select $1$ dictionary from $3$ is $\binom{3}{1} = 3$

Vedclass · Accepted Answer

At least $1000$

Vedclass · Answer

(B) The number of ways to select $4$ novels from $6$ is $\binom{6}{4} = 15$.The number of ways to select $1$ dictionary from $3$ is $\binom{3}{1} = 3$.To arrange these $5$ items ($4$ novels and $1$ dictionary) such that the dictionary is always in the middle,we fix the dictionary in the middle position in $1$ way.The remaining $4$ novels can be arranged in the remaining $4$ positions in $4!$ ways.Total arrangements $= \binom{6}{4} 	imes \binom{3}{1} 	imes 4! = 15 	imes 3 	imes 24 = 1080$.Since $1080 \geq 1000$,the correct option is $B$.

Out of $6$ distinct novels and $3$ distinct dictionaries,$4$ novels and $1$ dictionary are to be selected and arranged in a row on a shelf such that the dictionary is always in the middle. What is the number of such arrangements?

Similar Questions

In the product $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^{100})$,when written in ascending powers of $x$,the highest exponent of $x$ is . . . . . . .

If $f(n) = n! (31-n)!$,where $n \in \{0, 1, 2, \ldots, 31\}$,then the minimum value of $f(n)$ is

If all the letters of the word $'GANGARAM'$ are arranged,then the number of words in which exactly two vowels are together but no two $'G'$ occur together is-

If $a_n = \sum_{r=0}^n \frac{1}{^nC_r}$,then $\sum_{r=0}^n \frac{r}{^nC_r}$ equals

In the prime factorization of $37! = 2^{\alpha_2} \cdot 3^{\alpha_3} \cdot 5^{\alpha_5} \cdots 37^{\alpha_{37}}$,the ratio $\alpha_3 : \alpha_5$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module