How many 3 digit numbers can be formed from 0 9 if repetitions are allowed?

You can put this solution on YOUR website!
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For case b) I have another solution and different answer.
    The two last digits must be "25", or "50" or "75"  (00 does not work, since repetitions are not allowed).


    If the last two digits are "25", then the first digit can be any of only 7 digits  1, 3, 4, 6, 7, 8, 9.


    If the last two digits are "50", then the first digit can be any of 8 digits  1, 2, 3, 4, 6, 7, 8, 9.


    If the last two digits are "75", then the first digit can be any of only 7 digits  1, 2, 3, 4, 6, 8, 9.


    In all, there are  7 + 8 + 7 = 22 opportunities for the first digit and, correspondingly, 22 three-digit numbers

    satisfying the condition.

The 1st digit must be greater than or equal to 5 for the number to be greater than 500. There are 5 possibilities (5, 6, 7, 8, 9).

The 2nd digit has no restriction on it. There are 10 possibilities (0-9).

The 3rd digit must be odd in order for the number to be odd. There are 5 possibilities (1, 3, 5, 7, 9).

If what you want are all possible three digit numbers with no repetition of the digits then you have 10 choices for the first digit, you have 9 choices for the 2nd digit, and you have 8 choices for the 3rd digit giving you 10x9x8 = 720 in all.

How many three digit numbers can be formed by using the digits from 0 to 9 if the digits Cannot 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 three digit numbers can be formed using the digits 1 to 9?

Solution : No. of ways of filling first place = 9
No. of ways of filling second place = 8
No. of ways of fillin third place = 7
Therefore, total numbers `= 89 xx 8 xx 7 504`.

How many permutations of 3 different digits are there chosen from the digits 0 to 9 inclusive?

How many permutations of three different digits are there, chosen from the ten digits 0 to 9 inclusive? A. 84.

How many 3-digit even numbers can be formed using 0 to 9 without repetition?

The answer is - there are 328 three-digit even integers, with no digits repeated.

How many 3 digit numbers can be formed using the digits 0-9 if repetitions of digits are allowed?

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?

How many 3 digit numbers can be formed using the digits 0 1 2 and 5 if repetition of digits is not allowed?

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 numbers can be made with 3 digits?

Hence, there are 900 three-digit numbers in total.

How many three digit numbers can be formed?

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

How do you calculate the number of possible combinations?

To calculate combinations, we will use the formula nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. To calculate a combination, you will need to calculate a factorial.

How many 3 digit numbers can be formed from the digits 1 2 3 4 and 5 if the digits are unique?

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. Let 3-digit number be XYZ.

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

The answer is 12.

How many 3 digit numbers can be formed using the digits 1 7?

[127] would be the answer.

How many three digit area codes can be made from the digits 0 through 9 if the first digit is not allowed to be a 0 and the digits are allowed to repeat?

1 Expert Answer

There are 10 digits 0 1 2 3 4 5 6 7 8 9 if we can't use 0 and 1 then there are only 8 choices for the first digit and only 2 choices for the second digit and 10 for the third so there are 8*2*10 =160 possibilities.

The number of American households that were unbanked last year dropped to its lowest level since 2009, a dip due in part to people opening accounts to receive financial assistance during the pandemic, a new report says.  

Roughly 4.5% of U.S. households – or 5.9 million – didn't have a checking or savings account with a bank or credit union in 2021, a record low, according to the Federal Deposit Insurance Corporation's most recent survey of unbanked and underbanked households. 

Roughly 45% of households that received a stimulus payment, jobless benefits or other government assistance after the start of the pandemic in March, 2020 said those funds helped compel them to open an account, according to the biennial report which has been conducted since 2009.

"Safe and affordable bank accounts provide a way to bring more Americans into the banking system and will continue to play an important role in advancing economic inclusion for all Americans,'' FDIC acting chairman Martin J. Gruenberg said in a statement.  

A lack of banking options delayed some households from getting federal payments aimed at helping the country weather the economic fallout from the COVID-19 health crisis.

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The FDIC initiated an educational campaign to get more Americans to open an account to enable the direct deposit of those funds. And banks such as Capital One and Ally Financial ended  overdraft and other fees that have been a key barrier to some Americans accessing the banking system. 

What does it mean to be unbanked?

A household is deemed unbanked when no one in the home has an account with a bank or credit union. That share of households has dropped by nearly half since 2009. And since 2011, when 8% of U.S. households were unbanked, the highest since the start of the survey, and the record low reached in 2021, roughly half of the drop was due to a shift in the financial circumstances of American households the FDIC says.

Who are the underbanked?

A bank manager helps a woman open up a new account.

Those who have a checking or savings account, but also use financial alternatives like check cashing services are considered underbanked. The underbanked represented 14% of U.S. households, or 18.7 million, last year.   

Why are people unbanked or underbanked?

Many of those who are unbanked say they can't afford to have an account because of the fees for insufficient funds and overdrafts that are tacked on when account balances fall short. Roughly 29% said fees or not having the required minimum balance were the primary reasons they didn't have a checking or savings account, as compared to 38% who cited those obstacles in 2019.

Are some groups more likely to be unbanked? 

The numbers of the unbanked were greater among households that included those who were working age and disabled, lower income, included a single mother, or were Black or Hispanic. Among white households for instance, 2% didn't have a bank account last year as compared to 11% and 9% of their Black and Hispanic counterparts.

Meanwhile, nearly 15% of households with a working age member who had a disability were unbanked compared to almost 4% of other households. And  nearly 16% of households with a single mother were unbanked as compared to about 2% of married couples who lacked an account. 

 "These gaps attest there's still a lot of opportunity to expand participation across the population in the banking system,'' Keith Ernst, Associate Director of Consumer Research and Examination Analytics at the FDIC, said during a media call about the report.            

Will the number of unbanked rise if the U.S. has a recession? 

Perhaps.

"During the last recession unbanked rates did indeed go up,'' Karyen Chu, chief of the Banking Research Section at the Center for Financial Research, said during the call. 

Additionally, last year, homes where the head of household was out of work were nearly five times more likely to not have a bank account as compared to those where the household head was employed.

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"To the extent that income goes down ... that has generally been associated with increases in unbanked rates,’’ Chu said. 

How many 3

If what you want are all possible three digit numbers with no repetition of the digits then you have 10 choices for the first digit, you have 9 choices for the 2nd digit, and you have 8 choices for the 3rd digit giving you 10x9x8 = 720 in all.

How many 3 number combinations are there in the numbers 0 9?

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.