Acombination lock uses three numbers between 1 and 78 with repetition, and they must be selected in the correct sequence. which of the five counting rules is used to find that number? how many different "combinations" are possible? is the name of "combination lock" appropriate? if not, what other name would be better?

  first number      and       second number       and     third number   =     Total

78 possibilities      x         78 possibilities           x     78 possibilities   =     234

Since "order" matters, this is a permutation.

So, this can be calculated using: ₇₈P₃

"Permutation lock" would be a more appropriate name.

Step-by-step explanation:

According to the problem, the lock uses three numbers between 1 and 78, that is, 77 elements in total, with repetition. To find the answer we have to use the definition that allow elements to repeat, which is:

; where is the total number of elements, and is the subgroup.

Replacing values, we have:

Therefore, there are 456,533 ways to use the lock.

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step-by-step explanation:

the correct answer option is (2x-3)(2x+3).

step-by-step explanation:

we are given the following expression and we are to factorize it:

we are to find such factors that when multiplied they give a result of 36 (9 x 4 = 36) and when added they give a result of 0.

the two such factors of 36 are negative six and postive six.

factorizing the given expression to get:

therefore, the correct factorization of is (2x-3)(2x+3).