The number of seven-digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

Question

The number of seven-digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is (A) 55 (B) 66 (C) 77 (D) 88

Tardigrade Admin · Accepted Answer

There are two possible cases Case I Five 1 's , one 2's, one 3's Number of numbers = (7!/5!)=42 Case II Four 1's, three 2's Number of numbers =(7!/4!3!)=35 ∴ Total number of numbers = 42 + 35 = 77

Q. The number of seven-digit integers, with sum of the digits equal to $10$ and formed by using the digits $1, 2$ and $3$ only, is

55

66

77

88

There are two possible cases
Case I Five $1^{'} s$ , one $2^{'} s$ , one $3^{'} s$
Number of numbers = $\frac{7 !}{5 !} = 42$
Case II Four $1^{'} s$ , three $2^{'} s$
Number of numbers = $\frac{7 !}{4 ! 3 !} = 35$
$∴$ Total number of numbers $= 42 + 35 = 77$

Q. The number of seven-digit integers, with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only, is

55

66

77

88

There are two possible cases Case I Five 1′s , one 2′s, one 3′s Number of numbers = 5!7!​=42 Case II Four 1′s, three 2′s Number of numbers =4!3!7!​=35 ∴ Total number of numbers =42+35=77

Q. The number of seven-digit integers, with sum of the digits equal to $10$ and formed by using the digits $1, 2$ and $3$ only, is

There are two possible cases
Case I Five $1^{'} s$ , one $2^{'} s$ , one $3^{'} s$
Number of numbers = $\frac{7 !}{5 !} = 42$
Case II Four $1^{'} s$ , three $2^{'} s$
Number of numbers = $\frac{7 !}{4 ! 3 !} = 35$
$∴$ Total number of numbers $= 42 + 35 = 77$