If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p , then 96 p is equal to ......

Step-by-step solution

Total possibilities to make a 6-digit number using only 1 and 8 Every position has 2 possibility either 1 or 8 2^6 Number divisible by 21 : 1. for divisibility by 3 A (sum of digits must be divisible by 3 ) (a) Number of 1 's = Number of 8 (b) Every digit is 1 (c) Every digit is 8. 2. for divisiblity by 7. |2 (last digit) - remaining númber =7 k k Z (a) all are 1 ~s 1 no (b) all are 8 s 1 no (c) 31 's and 38 ~s 6! 3!3! Total number divisible by 21 = number divisible by 3 and 7 aligned & = 6! 3!3! +2= ( 6 5 4 3! 3!