Number of five digit integers, with sum of the digits equal to 43 are

Question

Number of five digit integers, with sum of the digits equal to 43 are (A) 5 (B) 10 (C) 15 (D) 35

Tardigrade Admin · Accepted Answer

Case-1: Digits being used are 9,9,9,9,7 then the number of numbers formed =(5 !/4 !)=5 Case-2: Digits being used are 9,9,9,8,8 then the number of numbers formed =(5 !/3 ! 2 !)=10 ∴ Total number of such numbers are 10+5=15

Q. Number of five digit integers, with sum of the digits equal to 43 are

5

10

15

35

Case-1 : Digits being used are $9, 9, 9, 9, 7$ then the number of numbers formed $= \frac{5 !}{4 !} = 5$
Case-2 : Digits being used are $9, 9, 9, 8, 8$ then the number of numbers formed $= \frac{5 !}{3 ! 2 !} = 10$
$∴$ Total number of such numbers are $10 + 5 = 15$

Q. Number of five digit integers, with sum of the digits equal to 43 are

5

10

15

35

Case-1 : Digits being used are 9,9,9,9,7 then the number of numbers formed =4!5!​=5 Case-2 : Digits being used are 9,9,9,8,8 then the number of numbers formed =3!2!5!​=10 ∴ Total number of such numbers are 10+5=15

Case-1 : Digits being used are $9, 9, 9, 9, 7$ then the number of numbers formed $= \frac{5 !}{4 !} = 5$
Case-2 : Digits being used are $9, 9, 9, 8, 8$ then the number of numbers formed $= \frac{5 !}{3 ! 2 !} = 10$
$∴$ Total number of such numbers are $10 + 5 = 15$