Let S = {2, 3, 4, 5, 6, 7, 9}. How many different 3-digit numbers [with all digits different] from S can be made which are less than 500?
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NDA 02/2021: Maths Previous Year paper [Held On 14 Nov 2021]
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Answer [Detailed Solution Below]
Option 3 : 90
Free
Electric charges and coulomb's law [Basic]
10 Questions 10 Marks 10 Mins
Explanation:
We can solve this by filling the places according to the question
Step 1:
Since the Number should be less than 500
So, there are only 3 possibilities at 1 st place i.e. [2, 3, 4]
Step 2:
According to the question
Repetition is not allowed we have to fix one number [let 2] in the 1st place
Now, At 2nd place, there are a total of 6 ways [3, 4, 5, 6, 7, 9]
Step 3:
According to the question
Repetition is not allowed we have to fix one number [let 3] in the 2nd place
Now, At 3rd place, there are a total of 5 ways [4, 5, 6, 7, 9]
∴ The total number of different 3-digit numbers less than 500 = 3 × 6 × 5 = 90.
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How many even numbers less than 500 can be formed using the digits 1,2,3,4,5? Each digits may be used only once in any number.
#1
[2, 4, 12, 14, 24, 32, 34, 42, 52, 54, 124, 132, 134, 142, 152, 154, 214, 234, 254, 312, 314, 324, 342, 352, 354, 412, 432, 452] = 28 such numbers
#2
Well we know the digit must end in 2 or 4, because it has to be even. So 1, 3, 5, can't be the ending digit.
We've chosen 2 digits already [2 and 4], so there are 3 possibilities for the first digit. It can't be 5, because then it would exceed the count of 500, so 1, 3, or whatever's left of 2 and 4. And only 3 digits left for the middle digit. So that means 2 * 3 * 3 = 18 cases for 3 digit numbers..
Now what about 2 digit numbers? Having chosen 2 or 4, we have 4 other possibilities left. So 2 * 4 = 8 cases for 2-digit numbers.
We have to remember 1 digit numbers, 2 or 4, so 2 cases.
In the end: 18 + 8 + 2 = 28 numbers.
13 Online Users
Case $1$: The number is a single digit.
In this case, the only even numbers are $2$ and $4$, giving a total of $2$.
Case $2$: The number has exactly two digits.
In this case, the last digit must be either $2$ or $4$, and the first digit must be one of the other four digits allowed, giving a total of $2 \cdot 4=8$.
Case $3$: The number has exactly three digits.
In this case, the last digit must be either $2$ or $4$ and the middle digit must be one of the other four digits allowed.
If the middle digit is $5$, then the first digit must be one of the other three digits allowed, giving a total of $2 \cdot 3=6$.
If the middle digit is not $5$, then the first digit must be one of two digits other than the last two digits and $5$, giving a total of $2 \cdot 3 \cdot 2=12$.
So, there are $2+8+6+12=28$ even numbers less than $500$ with distinct digits $\in \{1,2,3,4,5\}$.