What are the total number of ways in which a
four digit number that is divisible by 2, can be
formed using the numerals 0, 1, 2, 3, and 4
without repetition?
Answers
Answer:
4
Step-by-step explanation:
Case1: Take 0 at the units place
Remaining 3 places can be filled from the remaining 4 digits in 4P3 = 4!/(4–3)!=4!/1! = 24
Case2: Take 2 at the units place
As we require a 4 digit number, so the first digit cannot be chosen as 0 (zero). So the first digit can be 1, 3 or 4 i.e. 3 choices for the first place and then second place also 3 choices (2 from above and a 0) and finally 2 choices for the third place. This gives us, in all 3*3*2=18 such numbers.
Case3: Take 4 at the units place
This case is similar to the above case2. So in this case also there would be 18 numbers.
Hence in all there are 24+18+18=60 numbers satisfying above criteria.
The total numbers of ways is 60.
GIVEN
Four digit number that is divisible by 2, can be
formed using the numerals 0, 1, 2, 3, and 4 without repetition.
TO FIND
Total number of ways.
SOLUTION
We can simply solve the above problem as follows;
It is given that the 4-digit number formed should be divisible by 2. This means the digit in unit's place can be 0,2 or 4.
If zero is in the unit's place;
Combination of remaining 3 digits place = ⁴C₃ = 4!/(4-3)!
= 4!/1!
= 24.
If 2 is at unit's place.
Zero cannot be at thousandth's place.
The first digit can be 1, 3 or 4.
Therefore there are 3 choices for first digit.
3 choices for second place.
And,
2 choices for third place.
The combinations are- 3×3×2 = 18 combinations.
4 is at unit's place.
Combinations of numbers will be = 18.
Total combinations = 24 + 18 + 18 = 60 Numbers.
Hence, The total numbers of ways is 60.
#Spj2