A list is made of all the 3 digit numbers whose digits are 1, 2, 3, 4 or 5 with no repetition. If the sum of all the numbers is 10 N, then find the value of N.

Question

A list is made of all the 3 digit numbers whose digits are 1, 2, 3, 4 or 5 with no repetition. If the sum of all the numbers is 10 N, then find the value of N. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Total 3 digit numbers = 5 C 3 × 3 !=60 Sum of each column =12[1+2+3+4+5]=180 Hence sum =1998 × 10 ⇒ N =1998 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a0df6de814377868c37dc54677f032f9-.png" /> Alternative: Digit 1, 2, 3, 4, 5 are in A.P. and hence sum of all the 3 digit numbers is (3 !/2)(123+543)=3 × 666=1998

Q. A list is made of all the 3 digit numbers whose digits are 1, 2, 3, 4 or 5 with no repetition. If the sum of all the numbers is 10 N, then find the value of N.

Total 3 digit numbers $=^{5} C_{3} \times 3! = 60$
Sum of each column $= 12 [1 + 2 + 3 + 4 + 5] = 180$
Hence sum $= 1998 \times 10 \Rightarrow N = 1998$

Alternative : Digit 1, 2, 3, 4, 5 are in A.P. and
hence sum of all the 3 digit numbers is $\frac{3 !}{2} (123 + 543) = 3 \times 666 = 1998$

Q. A list is made of all the 3 digit numbers whose digits are 1, 2, 3, 4 or 5 with no repetition. If the sum of all the numbers is 10 N, then find the value of N.

Total 3 digit numbers =5C3​×3!=60 Sum of each column =12[1+2+3+4+5]=180 Hence sum =1998×10⇒N=1998 Alternative : Digit 1, 2, 3, 4, 5 are in A.P. and hence sum of all the 3 digit numbers is 23!​(123+543)=3×666=1998

Total 3 digit numbers $=^{5} C_{3} \times 3! = 60$
Sum of each column $= 12 [1 + 2 + 3 + 4 + 5] = 180$
Hence sum $= 1998 \times 10 \Rightarrow N = 1998$

Alternative : Digit 1, 2, 3, 4, 5 are in A.P. and
hence sum of all the 3 digit numbers is $\frac{3 !}{2} (123 + 543) = 3 \times 666 = 1998$