What number can you make using 3 5 7 use a digit only once?

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  • Given 3 digit numbers that can be formed using the digits from 1 to 9.Required number of 3-digit numbers= arranging 3 digits with the total number of 9 digits.= 9P3=(9×8×7)=504.Hence, the required number of numbers =504.
  • Problem 1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that – 
  • (i) repetition of the digits is allowed?
  • (ii) repetition of the digits is not allowed? 
  • Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
  • Problem 3: How many 4-letter code can be formed using the first 10 letters of the English alphabet if no letter can be repeated? 
  • Problem 4: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
  • Problem 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
  • Problem 6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
  • How many 3 digit numbers can be formed using the digits 1 3 5 7 9 where we are allowed to repeat the digits?
  • How many 3 digit numbers can be formed from digits 2 3 5 6 7 and 9 each digit used only once?
  • How many numbers can be formed using the digits 1 3 4 5 6 8 and 9 if I no repetitions are allowed II repetitions are allowed?
  • How many three digit multiples of can be written using numbers 1 3 5 9 if all digits are different?

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Solution

Given 3 digit numbers that can be formed using the digits from 1 to 9.Required number of 3-digit numbers= arranging 3 digits with the total number of 9 digits.= 9P3=(9×8×7)=504.Hence, the required number of numbers =504.

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How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 5, 6, 7, if no digits is repeated?

Q. How many 3-digit number can the formed by using the digit 1,2,3,4,5,6,7,8 if no digit is repeated?

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Q. How many 3 digit even numbers can be formed using the digits 3, 5, 7, 8, 9, if the digits are not repeated?

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Problem 1: How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that – 

(i) repetition of the digits is allowed?

Solution: 

Answer: 125.

Method:

Here, Total number of digits = 5

Let 3-digit number be XYZ.

Now the number of digits available for X = 5,

As repetition is allowed,

So the number of digits available for Y and Z will also be 5 (each).

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

(ii) repetition of the digits is not allowed? 

Solution:

Answer: 60.

Method:

Here, Total number of digits = 5

Let 3-digit number be XYZ.

Now the number of digits available for X = 5,

As repetition is not allowed,

So the number of digits available for Y = 4 (As one digit has already been chosen at X),

Similarly, the number of digits available for Z = 3.

Thus, The total number of 3-digit numbers that can be formed = 5×4×3 = 60.

Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Solution:

Answer: 108.

Method:

Here, Total number of digits = 6

Let 3-digit number be XYZ.

Now, as the number should be even so the digits at unit place must be even, so number of digits available for Z = 3 (As 2,4,6 are even digits here),

As the repetition is allowed,

So the number of digits available for X = 6,

Similarly, the number of digits available for Y = 6.

Thus, The total number of 3-digit even numbers that can be formed = 6×6×3 = 108.

Problem 3: How many 4-letter code can be formed using the first 10 letters of the English alphabet if no letter can be repeated? 

Solution:

Answer: 5040

Method:

Here, Total number of letters = 10

Let the 4-letter code be 1234.

Now, the number of letters available for 1st place = 10,

As repetition is not allowed,

So the number of letters possible at 2nd place = 9 (As one letter has already been chosen at 1st place),

Similarly, the number of letters available for 3rd place = 8,

and the number of letters available for 4th place = 7.

Thus, The total number of 4-letter code that can be formed = 10×9×8×7 = 5040.

Problem 4: How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Solution:

Answer: 336

Method:

Here, Total number of digits = 10 (from 0 to 9)

Let 5-digit number be ABCDE.

Now, As the number should start from  67 so the number of possible digits at A and B = 1 (each),

As repetition is not allowed,

So the number of digits available for C = 8 ( As 2 digits have already been chosen at A and B),

Similarly, the number of digits available for D = 7,

and the number of digits available for E = 6.

Thus, The total number of 5-digit telephone numbers that can be formed = 1×1×8×7×6 = 336.

Problem 5: A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Solution:

Answer: 8

Method:

We know that, the possible outcome after tossing a coin is either head or tail (2 outcomes),

Here, a coin is tossed 3 times and outcomes are recorded after each toss,

Thus, the total number of outcomes = 2×2×2 = 8.

Problem 6: Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Solution:

Answer: 20.

Method:

Here, Total number of flags = 5

As each signal requires 2 flag and signals should be different so repetition will not be allowed,

So, the number of flags possible for the upper place = 5,

and the number of flags possible for the lower place = 4.

Thus, the total number of different signals that can be generated = 5×4 = 20.

How many 3 digit numbers can be formed using the digits 1 3 5 7 9 where we are allowed to repeat the digits?

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 3 digit numbers can be formed from digits 2 3 5 6 7 and 9 each digit used only once?

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

How many numbers can be formed using the digits 1 3 4 5 6 8 and 9 if I no repetitions are allowed II repetitions are allowed?

Hence, 64 numbers can be formed.

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

The answer is 12.

What number can you make using 3 5and 7?

So, a total of 6 distinct numbers can be formed using these digits.

How many possible two digit numbers can be formed using the digits 3/5 and 7?

⇒ Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3. ∴ 9 possible two-digit numbers can be formed.

How many 3

UPLOAD PHOTO AND GET THE ANSWER NOW! Solution : 5, 0, 7 are the three-digit (given). The possible 3-digit numbers are: 507, 705, 750, 570.

How many numbers can be formed using the digits 1 3 5 and 7 provided no digit occurs twice?

=64. Was this answer helpful?