Then P10 can be filled up by any one of 1,2,3 ……. 9 in 9 different ways . ⇒n10=9 . After filling up P10 in any one way, 9 digits (including 0) remain so P9 can be filled up in 9 ways.

Hence, any of the 10 digits(0,1,2,3,4,5,6,7,8,9) can be placed at each place of the 8 digit number….Permutation and Combination #74.

74. How many 8 digits mobile numbers can be formed if at least one of their digits is repeated and 0 can also start the mobile number?
A. 107	B. 108 – 10P7
C. 108 – 10P8	D. 108

That exhausts the possibilities, so there are 120+300=420 even four digit numbers that can be formed using the digits 0,1,2,3,4,5,6. total required numbers = case 1 + case 2 = 120+300 = 420. Without considering different cases.

Without repetition counting the number of 3digit numbers using digits from {0,1,2,3,4,5}: Count the number of 3-digit strings whose last digit is even. This gives 60−8=52 3-digit even numbers using digits from {0,1,2,3,4,5} without repetition.

So, number of ways in which three digit even numbers can be formed from the given digits is 6×6×3=108.

Answer: The answer is 60.

So the total ways of Numbers starting with 2, in a 3 digit even number possible is 3P1×1=3. which means that ways of 3 digit even numbers more than 300 is 24−9=15; So total 48+48+15=111. 111 even numbers greater than 300 can be formed with the digits 1,2,3,4,5 if repetition of digits in a number is not allowed.

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

Total 4-digit even numbers: 480−60=420 By the way, it is number of choices, not probability. Consider cases: The last digit is 0. Then you have six choices for the first digit, five choices for the second digit, and four choices for the third digit, giving 6⋅5⋅4⋅1=120 four digit even numbers that end in 0.

Answer: 9998 is the right answer.

We can only use 5 numbers for the fourth digit. The number of 4 digit even numbers than can be formed using 0 to 9 is 504 * 5 = 2520.

In response to the question, there are 5*4*3*2 = 120 distinct permutations of 4 digit even numbers possible to be formed from the digits 1,3,5,6,8,9.

1000

1001

256

4 – digit numbers = 7502, 7205, 7052, 7025, 7520, 7250, 5702, 5207, 5270,5720,5072,5027, 2057, 2075, 2507, 2705,2750, 2570. These are the 18 numbers you can make with 7,5,0 and 2, using only once in a number.

So, the unit place can be filled by 7 ways. So, the required number of ways in which four digit numbers can be formed from the given digits is 9×9×8×7=4536.

What are all the possible combinations of 1234? Originally Answered: How many ways can you arrange 1234? Or: 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 4321.

Why Limit The Combinations To Only 7?

Characters	Combinations
4	24
5	120
6	720
7	5,040

The number of permutations of n objects taken r at a time is determined by the following formula: P(n,r)=n! (n−r)! A code have 4 digits in a specific order, the digits are between 0-9.

8068

Researchers at the data analysis firm Data Genetics have found that the three most popular combinations—“1234,” “1111,” and “0000”—account for close to 20 percent of all four-digit passwords.

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. 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.

The number of 5-digit combinations is 10 5=100,000. So, one more than 99,999.

127

120 possibilities

For any combination of items, each item is either included or not included in the combo. That means each item has 2 possibilities for every combination. For 6 items, that would make the number of combinations = 2^6 = 64.

27 unique possibilities

720 different permutations