Question

The sum of all the possible numbers of 4
digits formed by digits 3,5,5, and 6 using
each digit once is​

Answer 1

Answer:

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*a

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*aSame is for c

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*aSame is for c--> Sum of all values of c's places = 3333*c

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*aSame is for c--> Sum of all values of c's places = 3333*cNote that, b occurs 6 times each in thousands, hundreds, tens and units places

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*aSame is for c--> Sum of all values of c's places = 3333*cNote that, b occurs 6 times each in thousands, hundreds, tens and units places--> Sum of all values of b's places = 6*1000a + 6*100a + 6*10a + 6*a = 6666*a

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*aSame is for c--> Sum of all values of c's places = 3333*cNote that, b occurs 6 times each in thousands, hundreds, tens and units places--> Sum of all values of b's places = 6*1000a + 6*100a + 6*10a + 6*a = 6666*a--> Sum of all digits places = 3333*a + 6666*b + 3333*c

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*aSame is for c--> Sum of all values of c's places = 3333*cNote that, b occurs 6 times each in thousands, hundreds, tens and units places--> Sum of all values of b's places = 6*1000a + 6*100a + 6*10a + 6*a = 6666*a--> Sum of all digits places = 3333*a + 6666*b + 3333*c= 3333*(a + 2b + c)

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*aSame is for c--> Sum of all values of c's places = 3333*cNote that, b occurs 6 times each in thousands, hundreds, tens and units places--> Sum of all values of b's places = 6*1000a + 6*100a + 6*10a + 6*a = 6666*a--> Sum of all digits places = 3333*a + 6666*b + 3333*c= 3333*(a + 2b + c)= 3333*(3 + 10 + 6)

Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*aSame is for c--> Sum of all values of c's places = 3333*cNote that, b occurs 6 times each in thousands, hundreds, tens and units places--> Sum of all values of b's places = 6*1000a + 6*100a + 6*10a + 6*a = 6666*a--> Sum of all digits places = 3333*a + 6666*b + 3333*c= 3333*(a + 2b + c)= 3333*(3 + 10 + 6)= 63327

Explanation:

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