How many 3 digit numbers can be formed using 1, 2, 3

The different 3-digit numbers which can be formed by using the digits 0, 2, 5 without repeating any digit in the number are 205, 250, 502 and 520. Therefore, four 3 digit numbers can be formed by using the digits 0, 2, 5.

How many 3 digit numbers can be formed if no repetition is allowed?

There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed. Note: We can also use the multiplication principle to answer this question.

How many 3 digit numbers can be formed from the digits 12345?

Thus, The total number of 3-digit numbers that can be formed = 5×5×5 = 125.

How many three digits number can be formed from the digits 0 1 2 3 and 4 assuming that 2 repetition of digits are not allowed?

Therefore 48 Three digit numbers are possible.

How many 3 digit even numbers are formed using the digits 0 1 2 9 if the repetition of digits is not allowed?

=328. Was this answer helpful?

Total 3 Digit Numbers If Repetition allowed & Not allowed - Permutations & Combinations Problems

How many even 3 digit numbers can be formed from the digits 1/2 5/6 and 9 without repeating any digits?

Correct Option: D

(4 - 2)! Total ways = 3 × 4 × 2 = 24 ways. {∵ Total available digits are 1, 2, 5, 6, 9.

How many 3 digit numbers can be made up from the even digits from 0 to 9 you can include numbers starting with a 0?

Therefore the number of three digit even numbers ending with 0 is 9(8) =72.

How many three digit numbers can be formed using the digits 0 1 3 5 6 7 If I repetition of digits is not allowed II repetition of digits is allowed?

Therefore, a total of 100 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 when repetition is allowed.

How many three digit numbers can be written from the digits 1 2 3 and 4 if repetition of digits is allowed?

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.

How many 3 digit odd numbers can be formed from the digits 123456 if the digits can be repeated?

Hence, by the fundamental principle of multiplication, the required number of odd numbers `= (3xx6xx6) = 108. `

How many 3 digit numbers can be formed from the digits 2 3 5 6 7 and 9 which are divisible by 5 and none of the digits are repeated?

∴ Required number of numbers = (1 x 5 x 4) = 20.

How many 3 digit numbers can be formed using 2 3 4 and 5 with none of the digits being repeated?

Answer: There are 24 three digits numbers.

How many three digit multiples of 3 can be written using numbers 1 3 5 9 of all digits are different?

Answer: 195, 531, 591, 135, 315, 351 etc.

How many three digit numbers can be formed from the six digits are 2 3 5 6 7 and 9 when repetitions of digits are not allowed?

a) There are six digits 2, 3, 5, 6, 7 and 9. Supposed repetition are not allowed. 120 three-digit numbers can be formed.

How many outcomes can we get if we are to select 3 numbers from 0 to 9 with repetition?

Thanks, Mike. Hi Mike. If order mattered, we would say select the first number (10 choices), then the second (9 choices) then the third (8 choices). So there would be 10 x 9 x 8 = 720 possible choices.

How many 3 digit numbers can be formed by using the digits 1 to 9 if digits can be repeated?

⇒So, the required number of ways in which three-digit numbers can be formed from the given digits is 9×8×7=504.

How many numbers can be formed with the digits 1,2 3 4 3 2 1 so that the odd digits always occupy the odd places?

Expert-verified answer

We can form 7 digits numbers.

How many 3 digit numbers can you make using the digits 1/2 and 3 without repetition?

Answer: 24

So the number of digits available for B = 3 (As one digit has already been chosen at A), Similarly, the number of digits available for C = 2.

Solution : Given, total different things=n
the number of permutations of n things taken one at a time `=.^(n)P_(1)=hn`, now if we taken two at a time (repetition is allowed), then first place can be filled by n ways and second place can again be filled in n ways.
`therefore`The number of permutations of n things takenn two at a time
`=.^(n)P_(1)xx.^(n)P_(1)=nxxn=n^(2)`
Similarly, the number of permutations of n things taken three at a time`=n^(3)`
The number of permutations of n things taken r at a
time `=n^(r)`. hence, the total number of permutations
`=n+n^(2)+n^(3)+ . . .+n^(r)`
`=(n(n^(r)-1))/((n-1))` [sum of r terms of a GP]

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.

How many 3

How many 3

How many 3

How many 3 digits numbers can be formed by choosing from the digits 1,2 3 and 4 if the digits may be repeated?