How many numbers divisible by 5 and lying bet | vedclass

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

Question

(C) To be divisible by $5$,the unit place must be occupied by the digit $5$. Since the number must lie between $3000$ and $4000$,the thousandth place

Vedclass · Accepted Answer

$^4P_2$

Vedclass · Answer

(C) To be divisible by $5$,the unit place must be occupied by the digit $5$. Since the number must lie between $3000$ and $4000$,the thousandth place must be occupied by the digit $3$. We have $4$ remaining digits ${1, 2, 4, 6}$ to fill the remaining $2$ positions (hundreds and tens). The number of ways to arrange these $2$ digits from the $4$ available is given by $^4P_2$. Thus,the total number of such numbers is $^4P_2 = 4 	imes 3 = 12$.

How many numbers divisible by $5$ and lying between $3000$ and $4000$ can be formed from the digits $1, 2, 3, 4, 5, 6$ (repetition is not allowed)?

Similar Questions

In how many ways can a mixed doubles tennis game be arranged from $7$ married couples if no husband and wife are in the same game?

Let $X = \{(a, b) \mid a, b \in \mathbb{Z}, 0 \leq a \leq 20, 0 \leq b \leq 15\}$. Then the number of rectangles with vertices in set $X$ is

$A$ polygon has $44$ diagonals. The number of its sides is

The sum of all absolute values of the differences of the numbers $1, 2, 3, \ldots, n$,taken two at a time,i.e.,$\sum \limits_{1 \leq j < i \leq n} |i-j|$ equals:

Let $T_n$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides. If $T_{n+1} - T_n = 21$,then $n$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module