Question

The digits from 1 to 9 are written on 9 cards. Fred has the digit 7, 2 and 4, George has the digit 6, 5 and 1 and Donald has 8, 3 and 9. Each of them uses come of the four basic operations: + (addition), -(subtraction), x(multiplication),÷ (division); and each of his own cards exactly once. Who cannot obtain 20 as a result? _______________________

Answer 1

Answer:

Solution :

There are .9P9=9! ways of arranging the given digits to form a nine digit number. So, total number of 9 digit numbers formed by the given digits =9!.

Out of these 9! Numbers only those numbers are divisble by 4 which have their last digits as even natural number and the numbers formed by their last two digits are divisible by 4.

The various possibilities of last two digits are :

12, 32, 52, 72, 92

24, 64, 84

16, 36, 56, 76, 96

28, 48, 68

This means that there are 16 ways of choosing the last two digits. Corresponding to each of these ways the remaining 7 digits can be arranged in 7! ways. Therefore, the total number of 9 digit numbers divisble by 4 is 16×7!

Hence, required probability =16×7!9!=29