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How many 8 numbers combinations can you make from the numbers 1 to 49?
nCr = n!/((n-r)!r!) → 49C8 = 49!/((49-8)!8!) = 49!/(41!8!) = 450,978,066 combinations.
How many 5 digit combinations can you make with 8 numbers?
WRONG ANSWER NB !!If repeated numbers are permitted the answer is 8x8x8x8x8=32768 (85)If the numbers can only be used once in each combination, the answer is 8x7x6x5x4= 6720, where ABCDE has [n!/ (n-r)!] or [8! / 3!] combinations for 8 numbers used 5 at a time.NONONONO THIS IS NOT THE RIGHT FORMULA IT SHD BE [n ! / (n-r)! x r! ] = ONLY 56 COMBINATIONS where each number can only be used once.8x7x6x5x4x3x2x1 / (3x2x1) ( 5x4x3x2x1) = 56from napoleon solo
If using numbers 1 thru 49 how many 6 digit combinations can be made?
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
How many seven digits number can be made from numbers 123456789 but no repititions are allowed?
181440 possible numbers. 9 choices for first digit, leaving 8 choices for the second digit for each of these choices for the first digit, leaving 7 choices for the third digit for each of these choices for the second digit, leaving ... 3 choices for the seventh digit for each of these choices for the sixth digit, giving 9 x 8 x 7 x ... x 3 = 181440 possible numbers. More generally, when there are n different items and r need to be selected from them in order it is called a permutation and the number of ways of doing this is: nPr = n!/(n-r)! where the exclamation mark means "factorial" which is the product of all numbers from 1 to the number, that is n! = n x (n-1) x (n-) x ... x 2 x 1. In this case, there are n=9 items and r=7 need to be selected giving: 9P3 = 9!/2! = 9 x 8 x ... x 3 = 18144
What 3 digit number multiplied by a 2 digit number equals a four digit number and you must use all numbers 1 through 9?
I think you r cheating Mr.Cheater
How many 8 numbers combinations can you make from the numbers 1 to 49?
nCr = n!/((n-r)!r!) → 49C8 = 49!/((49-8)!8!) = 49!/(41!8!) = 450,978,066 combinations.
How many 5 digit combinations can you make with 8 numbers?
WRONG ANSWER NB !!If repeated numbers are permitted the answer is 8x8x8x8x8=32768 (85)If the numbers can only be used once in each combination, the answer is 8x7x6x5x4= 6720, where ABCDE has [n!/ (n-r)!] or [8! / 3!] combinations for 8 numbers used 5 at a time.NONONONO THIS IS NOT THE RIGHT FORMULA IT SHD BE [n ! / (n-r)! x r! ] = ONLY 56 COMBINATIONS where each number can only be used once.8x7x6x5x4x3x2x1 / (3x2x1) ( 5x4x3x2x1) = 56from napoleon solo
If using numbers 1 thru 49 how many 6 digit combinations can be made?
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
How many 6-digit numbers can be formed using the digits 23456 and 8 with repition?
Order does matter in this case so you need to use the formula for a Permutation with repetition allowed. This would be n^r where n=6 different numbers and r=6 possible places for each number to go when forming the full 6-digit number. 6^6 = 46,656 possible digits
How many different two digit numbers can you form using the digits 1 2 5 7 8 and 9 without repetition?
36 two digit numbers can be formed...(:From Rafaelrz: The question can be stated as;how many permutations of two different digits can beobtained from a set of six different digits ?Answer:nPr equals n!/(n-r) ...... for n = 6, r = 26P2 equals 6!/(6-2)! equals 30 Permutations.
How many seven digits number can be made from numbers 123456789 but no repititions are allowed?
181440 possible numbers. 9 choices for first digit, leaving 8 choices for the second digit for each of these choices for the first digit, leaving 7 choices for the third digit for each of these choices for the second digit, leaving ... 3 choices for the seventh digit for each of these choices for the sixth digit, giving 9 x 8 x 7 x ... x 3 = 181440 possible numbers. More generally, when there are n different items and r need to be selected from them in order it is called a permutation and the number of ways of doing this is: nPr = n!/(n-r)! where the exclamation mark means "factorial" which is the product of all numbers from 1 to the number, that is n! = n x (n-1) x (n-) x ... x 2 x 1. In this case, there are n=9 items and r=7 need to be selected giving: 9P3 = 9!/2! = 9 x 8 x ... x 3 = 18144
How do you multiply 2 digit by 2 digit using distributive property?
If PQ and RS are two 2-digit numbers, then PQ * RS = 100*P*R + 10*P*S + 10*Q*R + Q*S
What 3 digit number multiplied by a 2 digit number equals a four digit number and you must use all numbers 1 through 9?
I think you r cheating Mr.Cheater
How many 9 digit numbers r there?
Working with whole numbers only . . .- The smallest one is 100,000,000 .- The largest one is 999,999,999 .That's (all the counting numbers up to 999,999,999) minus (the first 99,999,999 of them) =900,000,000 numbers (nine hundred million)
How many combinations of 8 numbers are possible from 10 numbers?
In general, the number of combinations of n things taken r at a time isnCr = n!/[(n - r)!r!]Thus, we have:10C8= 10!/[(10 - 8)!8!]= 10!/(2!!8!)= (10 x 9 x 8!)/(8! x 2 x 1)= (10 x 9)/2= 5 x 9= 45
What r 3 numbers that have 2 4 and 8 as factors?
How about: 8 16 and 24
What is the highest 5 digit common multiple of 5 and 8?
Answer: Highest 5-digit anwer: 99999 5*8=40 99999/40 = 2499 r 39 99999-39=99960 Answer: 99960
