A combination lock has 5 different numbers. If each number can only be used ONCE, how many different combinations are possible?

1 answer:

5 0

Hello,

The first digit may be 0,1,2..,9 thus 10 possibilities
The second digit may be 0,1,2,3...,9 but not equal to the first thus 9 possibilities
The 3th digit may be 0,1,2..,9 but not the first and not the 2th =>8 possibiliteis
...
In Total, 10*9*8*7*6 combinations.

You might be interested in

Let g(x)=2x and h(x)=x^2+4 find the value. (hog)(a)

lisov135 [29]

In short, (h o g)(a) is just h( g(a) ).

so what we can do is simply get g(a) first and then plug that in h(x).

$\bf \begin{cases}
g(x)&=2x\\
h(x)&=x^2+4\\
(h\circ g)(a)&=h(~~g(a)~~)
\end{cases}
\\\\\\
g(a)=2(a)\implies g(a)=2a
\\\\\\
h(~~g(a)~~)\implies h(~~2a~~)=(2a)^2+4
\\\\\\
h(~~2a~~)=(2^2a^2)+4\implies h(~~2a~~)=4a^2+4$

8 0

4 years ago

The sum of the digits in a two digit number is 5. If 27 is subtracted from the number then places of the digits are interchanged

EleoNora [17]

Please find the attached photograph for your answer