An eight-digit number divisible by $9$ is to be formed using digits from $0$ to $9$ without repeating the digits. The number of ways in which this can be done is: (in $(7!)$)

  • A
    $72$
  • B
    $18$
  • C
    $40$
  • D
    $36$

Explore More

Similar Questions

The number of different words that can be formed from the letters of the word $SUCCESS$ in which the two $C$ are together but no two $S$ are together are:

The number of natural numbers lying between $1012$ and $23421$ that can be formed using the digits $2, 3, 4, 5, 6$ (repetition of digits is not allowed) and divisible by $55$ is $....$

Let $S_1 = \{(i, j, k) : i, j, k \in \{1, 2, \ldots, 10\}\}$,$S_2 = \{(i, j) : 1 \leq i < j + 2 \leq 10, i, j \in \{1, 2, \ldots, 10\}\}$,$S_3 = \{(i, j, k, l) : 1 \leq i < j < k < l, i, j, k, l \in \{1, 2, \ldots, 10\}\}$,$S_4 = \{(i, j, k, l) : i, j, k \text{ and } l \text{ are distinct elements in } \{1, 2, \ldots, 10\}\}$. If the total number of elements in the set $S_r$ is $n_r$ for $r = 1, 2, 3, 4$,then which of the following statements is (are) $TRUE$?
$(A) n_1 = 1000$
$(B) n_2 = 44$
$(C) n_3 = 220$
$(D) \frac{n_4}{12} = 420$

$A$ debate club consists of $6$ girls and $4$ boys. $A$ team of $4$ members is to be selected from this club,including the selection of a captain (from among these $4$ members) for the team. If the team has to include at most one boy,then the number of ways of selecting the team is

The number of ways in which $3$ children can distribute $10$ tickets out of $15$ consecutively numbered tickets among themselves such that they get consecutive blocks of $5, 3$ and $2$ tickets is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo