A password with 7 digits

Question

How passwords with $7$ digits can be when $8$ is the first digit, and all the passwords should be from the numbers : $7,8,8,8,9,9,9$

I think the answer is $5^6$ because we have 5 options starting from the second place. Am I right ?

Answer 1

The number of permutations of $n$ things when $a_1$ are of one kind, $a_2$ of another kind, and so on till some $a_k$ is $$n!\over a_1!a_2!\cdots a_k!$$

You have already fixed the first place with $8$, so for six places we have left one $7$, two $8$s, and three $9$s. This becomes

$${6!\over1!2!3!}=60$$

Answer 2

Since the question descrbe the passwords must be $7,8,8,8,9,9,9$ rather than $7,8,9$, I shall assume that you have to use $7$ exactly $1$ time, $8$ exactly $3$ times, and $9$ exactly $3$ times.

The first position has been fixed. Hence for the remaining $6$ numbers, we have to use $7$ exactly $1$ time, $8$ exactly $2$ times, and $9$ exactly $3$ times.

Hence it should be $$\frac{6!}{1!2!3!}$$

Remark for your attempt:

It is not clear to me what the $5$ options are.

Answer 3

Since the first digit is $8$, only $6$ digits left.

So you have two $8's$, three $9's $ and one $7's$.

To assign different numbers to each digit, consider assigning $9's$ first.

You are choosing $3$ out of $6$ digits to place all the $9's$, so there are $\displaystyle {6\choose3}$ combinations.

There are $3$ digits left and you may want to choose $2$ out of $3$ digits to place all the $8's$. So there are $\displaystyle {3\choose2}$ combinations.

There is only $1$ place left for the $7's$, so only 1 combination is possible.

So the total number of combination is $${6\choose3}\cdot{3\choose2}=60$$

In fact, there is no difference in the order of assignments in this case.