How many ways can three different numbers be selected from 10 numbers to open a keypad lock?
1 answer:
There are 720 ways through which 3 numbers can be selected from 10 numbers to open a keypad lock.
Given that there are 10 numbers are 3 numbers are required to open a keypad lock.
Permutations are the number of ways in which objects can be selected without replacements to form subsets.
It is expressed as n=n!/(n-r)!.
Factorial of a number is the product of consecutive numbers less than the number.
Total numbers available=10
Numbers to be selected=3
Number of ways can be found by using permutations.
Number of ways =10
=10!/(10-3)!
=10!/7!
=10*9*8*7!/7!
=10*9*8
=720 ways.
Hence there are 720 ways through which 3 numbers can be selected from 10 numbers.
Learn more about permutations at brainly.com/question/1216161
#SPJ1
You might be interested in
answer is -1 < x < 5
-|x - 2|+ 9 > 6
Rearrange the terms
-|x - 2| > 6 - 9
-|x - 2| > - 3
then divide both sides of the inequality by the co- efficient of variable
|x - 2| < 3
convert the absolute inequality to standard inequality
-3 <x - 2 < 3
separate compound inequalities into system of inequality
{x - 2}> -3
{x - 2 < 3}
Rearrange variable to the left side of the equation
x > -3 + 2
calculate the sum or difference
x > -1
x -2 < 3
Rearrange variable to the left side of the equation
x < 3 + 2
calculate the sum or difference
x < 5
x > -1 and x < 5
Find intersection
-1 < x < 5
Equations and inequalities are both mathematical sentences formed by relating two expressions to each other. In an equation, the two expressions are deemed equal which is shown by the symbol =. Where as in an inequality, the two expressions are not necessarily equal which is indicated by the symbols: >, <, ≤ or ≥.
learn more about inequality equation here :https://brainly.in/question/15934172
SPJ9
