• 072920 77839
  • info@smartachievers.in
Menu
Back
  • JEE ADVANCE
  • JEE MAIN
  • NEET
  • BITSAT
  • CBSE BOARD
  • UP BOARD
  • CUET
JEE ADVANCE
  • CLASS XI - XII
  • CRASH COURSE
CRASH COURSE
JEE MAIN
  • CLASS XI - XII
  • CRASH COURSE
CRASH COURSE
NEET
  • CLASS XI - XII
  • CRASH COURSE
BITSAT
  • CLASS XI - XII
CBSE BOARD
  • CLASS IX
  • CLASS X
  • CLASS XI
  • CLASS XII
  • CLASS VI
  • CLASS VII
  • CLASS VIII
UP BOARD
  • CLASS IX
  • CLASS X
  • CLASS XI
  • CLASS XII
CUET
  • CRASH COURSE
Courses


Solution


Q.

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is          [2025]
1 4608  
2 4607  
3 5719  
4 5720  

Ans.

(2)

Given digits are 0, 1, 2, 3, 4, 5, 6, 7.

First place can be filled in 3ways, i.e., (5, 6, 7)

For 1st place as 5, last place can be 0, 1, 2, 3

∴   Total number of ways = 1×512×4 = 2048

For 1st place as 6, last place can be 0, 1, 2

∴   Total number of ways = 1×512×3 = 1536

For 1st place as 7, last place can be 0, 1

Total number of ways = 1×512×2 = 1024

∴   Total number of ways = 2048 + 1536 + 1024 = 4608

So, 50000 is not included i.e., 4608 – 1 = 4607