How many even three digit numbers can be formed from the digits 1, 2, 5, 6 and 9

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hello friends here we have a question that how many even number of 4 digit can be formed with the digit even number we have to form with the digit 0 1 2 3 4 5 and 6 and no digit been used more than once that is a petition is not allowed write the digit 5 year is 0 1 2 3 4 5 and 6 how many digit we have 1234567 7 right now I want even digit 4 digit number that is 1234 now the even digit can be formed with 0 2 4 and 6 digit number so we are here we have zero right so here we will get a two cases

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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?
• (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 4 digit even code if you need to arrange 0 9 digits?
• How many 4 digit even numbers can be formed using the digits 0 1 2 3 4 5 6 A when digits can be repeated B when digits are not repeated?
• How many 4 digit even numbers can be formed?
• How many 4 digit even numbers are possible by using the digits 0 1 2 3 5 only once?

that is case one case one when we have at ones place we will have the choices that is to 4 and 6 not zero right so the ones place canfield in three ways slide and hear the given is no digit is repeated that mean once the number is selected here the remaining choices we have for the other digit that is we have 7 Number So here this thousand place can be filled with five ways right because 0 want fill send if 0 will come here then it won't be a four digit number it will be a three digit number right to the remaining 7 from the seven we once we fix a number we will have 6 choices the main and from the 62

This ones will be the zero so we will neglect s0 and we fixed we will get 5 number 5 number we can place at this place of the remaining here 0 will be also counted so here we will have five possible digit can be filled in display and here we have three number will be formed that is 5 into 5 into 4 into three that is equal to 300 right now case to case to that is that is 1234 right here we fix zero right so here we fix it is 1 number only two former even number right now the choices we have for this place will be 6 right here we will have 5

and here we will have 4 choices so this will come 6 into 5 into 4 into one that is equal to 120 night so we will add both cases the number we will get that is 0 24 25 number of even number we will get with this digit is 420 thank you

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),

How many 3

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.

How many 3

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

How many 3 digits number can be formed from the digits 12345?

∴ Total number of 3-digit numbers = 3×4×5=60.

How many 3

so 60(ans.)