dearteassociazione.org->Customizable Word problem Solvers ->Numbers-> SOLUTION: How countless different four-digit numbers can be created using the digits 1, 1, 9, and 9? var visible_logon_form_ = false;Log in or register.Username: Password: it is registered in one straightforward step!.Reset her password if girlfriend forgot it."; return false; } "> log in On Ad: over 600 dearteassociazione.org Word troubles at edhelper.com


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Click here to check out ALL troubles on numbers Word ProblemsQuestion 255054: How many different four-digit numbers have the right to be created using the number 1, 1, 9, and 9? found 2 options by CharlesG2, Theo:Answer through CharlesG2(834)
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(Show Source): You deserve to put this equipment on your website! How many different four-digit numbers have the right to be formed using the number 1, 1, 9, and also 9?119919191991911991919911combination formula, or the number of ways to integrate k items from a set of n:there are 4 number so n=4there are just 2 various numbers we are actually selecting so k=2 answer by Theo(11673)
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(Show Source): You can put this systems on her website! have to be equal to (4!)/(2!*2!) = (4*3*2*1)/(2*1*2*1) = 6let"s see what they would be:119999111919919119919119this winds up being a permutation with few of the aspects being the same.formula because that a permutation is n!if x the the n are the same, then the formula becomes n!/x!if, in addition to x being the same, y are also the same, climate the formula i do not care n!/(x!*y!)in your problem, n was equal to 4.x was same to 2 since you had 2 ones.y was also equal come 2 due to the fact that you had 2 nines.n!/(x!*y^1) ended up being 4!/(2!*2!)to view what the distinction is, look at 3 letter (a,b,c)how plenty of ways have the right to they it is in formed?3! = 3*2*1 = 6 ways due to the fact that each the the letters is different.they are:abcacbbacbcacabcbatake the exact same 3 letters and make 2 of castle the same.let"s assume a and also b end up being the very same letter d.you would certainly then have actually ddchow many ways can they be formed.we have n = 3 and also x = 2 and our formula is n!/x! = 3!/2! = 3the variety of ways they deserve to be created is 3 as follows:ddcdcdcddplease save in mind that we are talking around permutations here, and not combinations.with permutations, stimulate is important. The same facets in a different order space a different set.with combinations, order is not important. Each set has to have actually different elements in it.take the set of abc.with combinations, this develops one set. The same 3 aspects can only be used once regardless of what bespeak they room in the set.with permutations, this creates 6 sets.

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The exact same 3 aspects can be supplied 6 times in a various order each time.the number of combinations possible would it is in 1.the numer the permutations feasible would be 6.
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