The digits 1, 2, 3, 4, 5, 6 ,7 8, 9 are written in random order to form a 9 digit number find the probability to be divisible by 11
Answers
Given : Digits from 1 to 9 to form 9 Digit Number
To Find : Probability that number is Divisible by 11
Solution :
Divisibility rule for 11
The Divisibility Rule of Eleven is that you must subtract and then add the digits in an alternating pattern from left to right
and resultant should be divisible by 11
The digits 1, 2, 3, 4, 5, 6 ,7 8, 9
A B C D E F G H I
A + B + C + D + E + F + G + H + I = 45
(A + C + E + G + I) - (B + D + F + H ) should be divisible by 11
(A + C + E + G + I) + (B + D + F + H ) = 45
=> one is odd and another is even hence 0 , -22 , 22 not possible
only possible -11 & 11
(A + C + E + G + I) + (B + D + F + H ) = 45
(A + C + E + G + I) - (B + D + F + H ) = 11 or - 11
=> B + D + F + H = 17 or 28
28 - (9 , 8 , 7 , 4) , ( 9 , 8 , 6 , 5)
17 - ( 9 , 5 , 2 , 1) , (9 , 4 , 3 , 1) , ( 8 , 6 , 2 , 1) , ( 8 , 5 , 3 , 1) , ( 8 , 4 , 3 , 2) , (7 , 6 , 3 , 1) , (7 , 5 , 4 , 1) , (7 , 5 , 3 , 2) , ( 6 , 5 , 4 , 2)
Total possible cases = 11
and 4 Digits can be arranged in 4! Ways
remaining 5 Digits can be arranged in 5! ways
9 Digits can be arranged in 9! ways
Hence Probability = 11 * 5! * 4 ! / 9 !
= 11 * 4 * 3 * 2/ (9 * 8 * 7 * 6)
= 11/126
11/126 is the probability that 9 digit number formed with The digits 1, 2, 3, 4, 5, 6 ,7 8, 9 are divisible by 11
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