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natali 33 [55]
3 years ago
6

A keycode must contain 2 letters and 3 numbers. The letters may be any letter of the alphabet. The numbers should be any number

from 0 to 9. How many different keycode combinations are there?
Mathematics
2 answers:
OverLord2011 [107]3 years ago
8 0
The English alphabet contains 26 letters (a, b, c, ...y, z). 

The digits from 0 to 9 are a total of 10.


A keycode contains 2 letters, and 3 numbers, for example:

AB 598;  MM 139;    NT 498; ...


So there are 26 possible choices for the first letter, which can combined with any of the 26 possible choices for the second letter, so there are a total of 

26*26=676 possible pairs of letters.


Similarly, the 10 possible choices for the first number, which can be combined with the 10 possible choices for the second number, and the 10 possible choices for the third number make a total of :

10*10*10=1,000 possible triples of numbers.


Any of the 676 possible pairs of letters can be combined with any of the possible 1,000 triples of numbers. This makes a total of 

676*1,000=676,000 keycodes.


Answer: 676,000
professor190 [17]3 years ago
7 0

Answer:

There is 676,000 combinations

Step-by-step explanation:

VOTE ME FOR BRAINILEST

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3 years ago
Nadia is ordering cheesecake at a restaurant, and the server tells her that she can have up to five toppings: caramel, whipped c
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Answer:

The probability that Nadia gets just caramel, butterscotch sauce, strawberries, and hot fudge is P =  1/32 = 0.03125

Step-by-step explanation:

There are up to 5 toppings, such that the toppings are:

caramel

whipped cream

butterscotch sauce

strawberries

hot fudge

We want to find the probability that,  If the server randomly chooses which toppings to add, she gets just caramel, butterscotch sauce, strawberries, and hot fudge.

First, we need to find the total number of possible combinations.

let's separate them in number of toppings.

0 toppins:

Here is one combination.

1 topping:

here we have one topping and 5 options, so there are 5 different combinations of 1 topping.

2 toppings.

Assuming that each topping can be used only once, for the first topping we have 5 options.

And for the second topping we have 4 options (because one is already used)

The total number of combinations is equal to the product between the number of options for each topping, so here we have:

c = 4*5 = 20 combinations.

But we are counting the permutations, which is equal to n! (where n is the number of toppings, in this case is n = 2), this means that we are differentiating in the case where the first topping is caramel and the second is whipped cream, and the case where the first topping is whipped cream and the second is caramel, to avoid this, we should divide by the number of permutations.

Then the number of different combinations is:

c' = 20/2! = 10

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for the second there are 4 options

for the third there are 3 options

the total number of different combinations is:

c' = (5*4*3)/(3!) = (5*4*3)/(3*2) = 10

4 toppings:

We can think of this as "the topping that we do not use", so there are only 5 possible toppings to not use, then there are 5 different combinations with 4 toppings.

5 toppings:

Similar to the first case, here is only one combination with 5 toppings.

So the total number of different combinations is:

C = 1 + 5 + 10 + 10 + 5 + 1 = 32

There are 32 different combinations.

And we want to find the probability of getting one particular combination (all of them have the same probability)

Then the probability is the quotient between one and the total number of different combinations.

p = 1/32

The probability that Nadia gets just caramel, butterscotch sauce, strawberries, and hot fudge is P =  1/32 = 0.03125

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\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}


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