a) How many five digits numbers divisible by 3 can be formed using the digits 0,1,2,3,4, 7 and 8 if each
digit is to be used atmost once.
Answers
Answer:
744
Step-by-step explanation:
To make the 5 digit number we have to remove two digits out of seven.
For being divisible by 3 the sum of digits of number must be divisible by 3.
The sum of all digits is 25. We have to remove two numbers in such a way that the sum will be divisible by 3, for example if we remove 0 and 1, the sum of digits will be 25-1=24 which is divisible by 3.
Thus the sets of two digits which can be removed are:
(0,1),(0,4),(0,7),(1,3),(2,8),(3,4),(3,7)
Now we have to consider two cases, when we remove 0 and when we not remove it. Because when we not remove 0 we have to make numbers suchthat 0 is not placed at leftmost place.
Case 1: When we remove 0 [(0,1),(0,4),(0,7)]
The number of 5-digit numbers formed using five distinct digits = 5!=120
The total number for this case= 3*120=360
Case 2: When we not remove 0 [(1,3),(2,8),(3,4),(3,7)]
The number formed using 5 digits =5!=120
The number in which 0 is at leftmost place=4!=24
The 5 -digit numbers formed =120-24=96
The total number formed in this case= 4*96=384
Thus the total 5-digit numbers formed =360+384=744