How many 3-digit numbers can be made using the digits 1,2,3,4,5 if the number must be less than 300?

Solution : For a number to be odd, we must have 1, 3 of 5 at the unit's palce. So, there are 3 ways of filling the unit's place.
Case (i) When the repetition of digits is not allowed:
In this case, after filling the unit's place, we may fill the ten's place by any of the remaining five digits. So, there are 5 ways of filling the ten's place.
Now, the hundred's place can be filled by any of the remaining 4 digits. So, there are 4 ways of filling the hundred's place.
So, by the fundamental principle of multiplication, the required number of odd numbers `=(3xx5xx4) = 60.`
Case (ii) When the repetition of digits is allowed:
Since the repetition of digits is allowed, so after filling the unit's place, we may fill the ten's place by any of the given six digits. So, there are 6 ways of filling the ten's place. Similarly, the hundred's palce can be filled by any of the given six digits. So, it can be filled in 6 ways.
Hence, by the fundamental principle of multiplication, the required number of odd numbers `= (3xx6xx6) = 108.`

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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1. 30
2. 49
3. 90
4. 147

Answer (Detailed Solution Below)

Option 3 : 90

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Electric charges and coulomb's law (Basic)

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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 (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 $3$ digit numbers can be formed by $1,2,3,4$, when the repetition of digits is allowed?

So basically, I attempted this question as-

There are 4 numbers and 3 places to put in the numbers: In the ones place, any 4 numbers can be put, so there are 4 choices in the ones place. Similarly for the tens and the hundreds place. So, the total choices are, by multiplication principle- $$4*4*4=64$$ And well and good, this was the answer.

But what if I reversed the method?

So I take some particular numbers, like $1,2,3$ and say that, well, $1$ can go in $3$ places, $2$ in $2$ places and $3$ in $1$ place, so by multiplication principle, there are $6$ ways of forming a $3$-digit number with $1,2,3$.

But there are $4$ different numbers. So the number of $3$-number combinations are- $(1,2,3)$,$(1,2,4)$,$(1,3,4)$,$(2,3,4)$. Each can be arranged in $6$ ways, so we get $24$ ways totally.

So why is my answer different here?

We interpret "number" as meaning a string of digits, so the problem asks us to find the number of five-digit strings taken from $\{0,1,3,4,5,7,9\}$ which contain at least one $0$, at least one $4$, and at least one $5$. It's possible for such a string to start with $0$. (This is contrary to the usual terminology in combinatorics, where a "number" cannot start with $0$.)

We will use the Principle of Inclusion and Exclusion. Without the restriction on the required digits, there are $N=7^5$ five-digit strings taken from the given set of digits.

Let's say a five-digit string has "Property $i$" if it does not contain the digit $i$, for $i\in \{0,4,5\}$, and define $S_j$ as the number of strings with $j$ of the properties, for $j = 1,2,3$. To compute $S_j$, note that the $j$ omitted digits can be chosen in $\binom{3}{j}$ ways, and then there are $(7-j)^5$ ways to arrange the remaining $7-j$ digits in a string of length $5$, so $$S_j = \binom{3}{j}(7-j)^5$$ By inclusion / exclusion, the number of five-digit strings with none of the properties, i.e. the number of strings with at least one each of $0,4$ and $5$, is $$N-S_1+S_2-S_3 = \boxed{1830}$$

If we adopt the convention that the number cannot start with $0$ then a solution by inclusion / exclusion is still possible, but it's more complicated.

How many 3

How many 3

How many 3

How many 3 digits numbers can be formed from the digits 1 2 3 4 and 5 Assuming that a repetitions of digits are allowed B repetitions of digits are not allowed?