MCQ
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:
- A$72 (7!)$
- B$18 (7!)$
- C$40(7!)$
- ✓$36(7!)$
Now sum of the digits from $0$ to $9$
$=0+1+2+3+4+5+6+7+8+9$
$=45$
Hence to from $8$ digits numbers which are divisible by $9$, a pair of digits either $0$ and $9,1$ and $8,2$ and $7,3$ and $6$ or $4$ and $5$ are not used.
| Digits which are not used to from $8$ digits number divisible by $9$ | Number of $8$ digits numbers which are divisible by $9$ |
| $0$ and $9$ | $8 \times 7!$ |
| $1$ and $8$ | $7 \times 7!$ |
| $2$ and $7$ | $7 \times 7!$ |
| $3$ and $6$ | $7 \times 7!$ |
| $4$ and $5$ | $7 \times 7!$ |
Hence total number of $8$ digits numbers which are divisible by $9$
$=8\times (7!)+7 \times (7!)+7\times (7!) +7 \times (7!) + 7\times (7!)$
$=36 \times (7!)$
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$1.$ The numbers $\left|A_1\right|,\left|A_2\right|, \ldots,\left|A_m\right|$ are distinct.
$2.$ $A_1, A_2, \ldots, A_m$ are pairwise disjoint.(Here $|A|$ donotes the number of elements in the set $A$ )Then, the maximum possible value of $m$ is