The number of seven digit integers,with sum of the digit equal to 10 and formed by using digits 1,2 & 3 only.
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Hi , why did you share this set ? It is empty , so it won't do anyone any good .
So before people get angry about it , at least please change it's name from "to solve" to "Empty Set" ,for example ..
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OK bro Done it! Change the name to "empty".
Done using generating functions!
a 1 + a 2 + a 3 + . . . + a 7 = 1 0 for a i s in [1,3]
( x + x 2 + x 3 ) 7 is our function and we have to find the co-efficient of x 1 0
x 7 ( ( 1 + x + x 2 ) 7 )
So, we just have to find co-efficient of x 3 in ( ( 1 + x + x 2 ) 7 )
( 1 − x 3 ) 7 ( ( 1 − x ) − 7 )
In the first, we will just get minimum of x 3 [except that one] and its co-efficient is − C ( 7 , 1 )
In the second, we will get the coefficient of x 3 is C ( 7 + 3 − 1 , 3 ) = C ( 9 , 3 )
Therefore, our answer becomes C ( 9 , 3 ) − C ( 7 , 1 ) = 7 7
My answer is same as that of Raghav Gupta , probably better presented here.
There are only 2 possible cases:
N 1 = 4 ! 3 ! 7 ! = 3 5
N 2 = 5 ! 7 ! = 4 2
Therefore, the total number of seven-digit numbers is:
N = N 1 + N 2 = 3 5 + 4 2 = 7 7
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There are 2 possible cases,
Case1- Five 1's,one 2's,one 3's No.=7!÷5!=42
Case 2- Four 1's,three 2's No.of numbers=7!÷4!3!=35
Total no. Of numbers= 42+35=77