How many 3-digit numbers can be formed from the digits 0, 1, 2, 3, 4,5 if repetition is allowed

Answer

Hint: Fundamental principle of counting: According to the fundamental principle of counting if a task can be done in “m” ways and another task can be done in “n” ways, then the number of ways in which both the tasks can be done in mn ways.Complete step by step solution:Case [1]: Repetition not allowed:The number of ways in which the ones place can be filled = 5 ways.The number of ways in which the tens place can be filled = 4 ways (because repetition is not allowed. So, the choice of one place cannot be used). The number of ways in which hundreds place can be filled = 3 ways.Hence according to the fundamental principle of counting the number of ways in which the three places can be filled to form a three-digit number = $5\times 4\times 3=60$Hence the number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 = 60 i.e 5! waysCase [2]: Repetition is allowed:The number of ways in which the ones place can be filled = 5 ways.The number of ways in which the tens place can be filled = 5 ways (because repetition is allowed. So, the choice of one's place can be used). The number of ways in which hundreds place can be filled = 5 ways.Hence according to the fundamental principle of counting the number of ways in which the three places can be filled to form a three-digit number = $5\times 5\times 5=125$ Hence the number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 = 125Note: The number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 when repetition is not allowed is equivalent to the number of 3 Letter permutations of 5 distinct letter = ${}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!}=\dfrac{5!}{2!}=\dfrac{120}{2}=60$ which is the same result as above.

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Answer:

Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digits) = 5*4*3*2*! = 5! = 120.

Text Solution

Solution : (i) When repetition of digits is allowed: <br> No. of ways of choosing firsy digits = 5 <br> No. of ways of choosing second digit = 5 <br> No. of ways of choosing third digit = 5 <br> Therefore, total possible numbers `= 5 xx 5 xx 5 = 125` <br> (ii) When repetition of digits is not allowed: <br> No. of ways of choosing first digit = 5 <br> No. of ways of choosing second digit = 4 <br> No. of ways of choosing thrid digit = 3 <br> Total possible numbers `= 5 xx 4 xx 3 = 60`.

You have 7 digits to choose from. After using one of them, you will only have 6 and after using another, you will only have 5. So this means that the number of possible combinations are 7*6*5 = 42*5 = 210

So there are 210 ways that a three digit number can be formed using the digits 0,1,2,3,4,5,6 and repetition is not allowed

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1) How many 3-digit numbers can be formed by using $0,1,2,3,4,5$ ? Using basics it would be $ 5 \times 5 \times4 = 100$

2) How many 3-digit numbers can be formed by $8,1,2,3,4,5$ which are even? Again using basics we get $ 4 \times 5 \times 3 =60$

3) Now I want to ask how many 3 digit numbers can be formed which are even using $0,1,2,3,4,5$?

No repetition is allowed in all above cases. Here I am not getting how to use basics when we need to apply both conditions of case 1 and case 2 (i.e when we need to take care of both things zero at hundredth place and even number at unit place) simultaneously .