How many positive integers n can we make with the digits 3, 3, 4, 5, 5, 6, 7, if the number n > 4, 000, 000?

Question

According to my study guide the answer to the exercise,

How many positive integers, (n), can we make with the digits 3, 3, 4, 5, 5, 6, 7, if the number n > 4, 000, 000, :

The total of numbers n > 4, 000, 000 is equal to the number of arrangements of 3, 3, 5, 5, 6, 7 of size 6; thus,= $\frac{6!}{2!2!} = 180$.

The total of numbers n > 5, 000, 000 is equal to the number of arrangements of 3, 3, 4, 5, 6, 7 of size 6; thus, $ \frac{6!}{2!}= 360$.

The total of numbers n > 6, 000, 000 is equal to the number of arrangements of 3, 3, 4, 5, 5, 7 of size 6; thus, $\frac{6!}{2!2!} = 180$

The total of numbers n > 7, 000, 000 is equal to the number of arrangements of 3, 3, 4, 5, 5, 6 of size 6; thus, $\frac{6!}{2!2!}=180$

The total is 3(180) + 360 = 900.

There are a couple of things I don't understand regarding this answer.

1. Why are we getting n > 4,000,000, n > 5,000,000, and so forth?
2. Why are we dropping the 4 from the set of digits when figuring out n > 4,000,000, the 5 from n > 5,000,000, and so on?

I thought I would need to use something like:

$\frac{n!}{(n-r)!}$ = $\frac{7!}{(7-2)!}$ to get the answer. I have (7-2)! because the number 3 < 4 and therefore cannot make a number > 4,000,000 but, when using all 7 digits the remaining numbers will create numbers > 7. For example, 4,556,733 > 4,000,000 and 5,567,334 > 4,000,000, but 3,345,567 < 4,000,000.

I hope I explained this properly. Thanks for any insight.

Tony

Answer 1

The solution you have breaks it into cases: A separate case is considered for $4$ being the leading digit, $5$ being the leading digit, and so forth all the way up to $7$. Here I offer this alternate solution:

In total, one can make $\frac{7!}{2! 2!}$ distinct numbers, disregarding the $n > 4,000,000$ condition. We can use complementary counting now -- take the total and subtract $n < 4,000,000$

In this case, $n < 4,000,000$ if and only if the leading digit is $3$. The number of such numbers with leading digit three is equal to the number of permutations of the digits $3, 4, 5, 5, 6, 7$, which is $\frac{6!}{2!}$.

This gives an answer of $\frac{7!}{ 2! 2!} - \frac{6!}{2} = 900$, agreeing with your study guide.