7 digits from the number 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in a random order (without repetation). The chance that the 7 digit number is divisible by 9 is :

Question

7 digits from the number 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in a random order (without repetation). The chance that the 7 digit number is divisible by 9 is : (A) 1/3 (B) 5/36 (C) 1/9 (D) 11/72

Tardigrade Admin · Accepted Answer

n ( S )= 9 C 7 ⋅ 7 ! n ( A )=4 ⋅ 7 ! (4 ⋅ 7 !/ 9 C 7 ⋅ 7 !)=(4/36) ∴ P ( A )=(4/36)=(1/9)

Q. 7 digits from the number $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a random order (without repetation). The chance that the 7 digit number is divisible by 9 is :

1/3

5/36

1/9

11/72

$n (S) =^{9} C_{7} \cdot 7!$
$n (A) = 4 \cdot 7!$
$\frac{4 \cdot 7 !}{^{9} C _{7} \cdot 7 !} = \frac{4}{36}$
$∴ P (A) = \frac{4}{36} = \frac{1}{9}$

Q. 7 digits from the number 1,2,3,4,5,6,7,8,9 are written in a random order (without repetation). The chance that the 7 digit number is divisible by 9 is :

1/3

5/36

1/9

11/72

n(S)=9C7​⋅7! n(A)=4⋅7! 9C7​⋅7!4⋅7!​=364​ ∴P(A)=364​=91​

Q. 7 digits from the number $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a random order (without repetation). The chance that the 7 digit number is divisible by 9 is :

$n (S) =^{9} C_{7} \cdot 7!$
$n (A) = 4 \cdot 7!$
$\frac{4 \cdot 7 !}{^{9} C _{7} \cdot 7 !} = \frac{4}{36}$
$∴ P (A) = \frac{4}{36} = \frac{1}{9}$