# Gyan Room - Modern Math - Concepts & Shortcuts

This thread is reserved for sharing concepts, short cuts and good questions from Modern Math topic.

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• Sum of all numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time (means repetition is not allowed)

The total number of numbers formed with the digits 2, 3, 4, 5 taken all at a time = 4! = 24
to find the sum of these 24 numbers we will find sum of digits at unit, ten, hundred and thousand places in all these numbers.
consider the digits in the unit's places in all these numbers
Each of the digit will occur 3! =6 times in the unit place
So total for the digits in unit's place in all the numbers =(2 + 3 + 4 + 5) * 6
so sum of all numbers =(2 + 3 + 4 + 5) * 3! * (10^0+10^1+10^2+10^3))
so (2 + 3 + 4 + 5) * 3! * (10^4-1)/9
= (2 + 3 + 4 + 5) * 3! * (1111)

Direct Formula:
Sum of numbers formed by n non zero digits = (sum of digits) * (n-1!) * (11111....n times)

When repetition allowed
total number of numbers in case of repetition = 4 * 4 * 4 * 4 = 4^4
So each digit will occur at 4^4/4 = 4^3 = 64 times
so (sum of digits) * 64 * (1111))

Direct Formula = sum of digits * n^n * (1111)

Find sum of all the numbers greater than 10000 formed by digits 0, 2, 4, 6, 8 (No digit is repeated in any number)

Sum = sum of numbers formed by using digits (0,2,4,6,8) - sum of numbers formed by using digits (2,4,6,8)
(0 + 2 + 4 + 6 + 8) * (5-1!) * (11111) - (2 + 4 + 6 + 8) * (4 - 1!) * (1111))

• The possible shortest routes in which we can travel from A to B (grid of m x n) are ( m + n )! / m!n!

The possible shortest routest in which we can travel from A to B where there is a shortcut are:

For example in grid of ( 6 x 4 )
From A to C in 4!/2!2! = 6 ways
From C to D = 1 way
From D to B = 4!/1!3! = 4 ways
Total ways = 6 x 1 x 4 = 24 ways

• Picking r people out of n people sitting in a circle such that no two are adjacent = (n-r+1)Cr - (n-r-1)C(r-2)
Picking r people out of n people sitting in a row such that no two are adjacent = (n - r + 1)Cr

@hemant_malhotra

• Credits @gaurav_sharma

There is a frog who could climb either 1 stair or 3 stairs in one shot. In how many ways he could reach at 10th stair ?

Fibonacci with a gap of 1 : 1, 1, 2, 3, 4, 6, 9, 13, 19, 28

Method 2 :
1 step in 1 way
2 steps in 1 way
3 steps in 2 ways
4 steps in 3 ways
5 steps in 4 ways
6 steps in 6 ways
7 steps in 9 ways
8 steps in 13 ways
9 steps in 19 ways
10 steps in 28 ways
So 28 should be the answer

Method 3 :

x + 3y = 10
(1 , 3 ) -> 4
(4 , 2 ) -> 6!/4!2! = 15
(7 , 1) -> 8!/7! = 8
(10 , 0) -> 1
TOTAL = 15 + 4 + 8 + 1 = 28

• Shortcut for Finding the Rank Of A Word

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