Why is coin tossing combination *without* repetition? As per the title, for a traditional statistics question like:
Find the probability of landing 7 heads in 10 throws of fair coin, P(head) = 0.5
No problems with this being a combination (as opposed to a permutation), but I cannot convince myself why it's the WITHOUT repetition sub category of combinations. Thus allowing one to use the Binomial Distribution via nCr. As opposed to using the n+r-1Cr distribution for WITH repetition.
What exactly does "without repetition" physically mean in this real life example?

Perhaps more generally than in my Comments, consider an urn with
ten marbles: five red and five blue. What is the practical
difference between sampling with replacement and sampling without replacement?
Suppose you choose five marbles out of ten, with replacement.
That is, you take out a marble, note its color, and put the marble
back in the urn. So, on each draw the urn has $5/10 = 0.5$ red marbles and $5/10 - 0.5$ blue marbles. Following the same logic is for
tossing a fair coin five times, we see that the probability of
drawing a red marble in exactly two of the five draws is ${5\choose 2}(1/2)^5 = 10/32 = 0.3125.$ [As before, the 'binomial coefficient' ${5\choose 2} = 10$ refers to the arrangements of two red and three marbles in sequence.]
By contrast, if I do not replace eacg=h of the five marbles as I draw them, then the contents of the urn are different on each draw. So, we have to say
that there are ${10\choose 5}$ ways to draw five marbles from among ten. The number of ways to choose exactly two red marbles out of five
is ${5\choose 2} = 10$ and the number of ways to choose exactly three blue
marbles out of five is ${5\choose 3} = 10.$ Then the number of ways to choose exactly 2 red marbles in five draws without replacement from
the urn is the 'hypergeometric' probability
$$\frac{{5\choose 2}{5\choose 3}}{{10\choose 5}} = \frac{10(10)}{252} = 0.3968254.$$
The bottom line is that the difference between (binomial)
sampling with replacement and (hypergeomatric) sampling
without replacement has given different probabilities
for getting exactly two red marbles in five draws from
the urn---binomial $0.3125$ and hypergeometric $0.3968254.$
In R:
choose(10, 5)
[1] 252
choose(5,2)
[1] 10
dhyper(2, 5,5, 5)  # without replacement
[1] 0.3968254
dbinom(2, 5, .5)   # with replacement
[1] 0.3125

The plot below shows the binomial and hypegeometric PDFs for
values $x = 0,1,2,3,4,5.$ Hypergeometric probabilities for $x=0$ and $x=5$ are quite small (almost too small to show on the graph). The hypergeometric distribution has smaller variance because options decrease as the marbles in the urn
disappear.

R code for the figure:
k = 0:5
pdf.b = dbinom(k, 5, .5)
pdf.h = dhyper(k, 5,5, 5)
hdr = "PDFs of Binomial (blue) and Hypergeometric Distributions" 
plot(k-.04, pdf.b, type="h", ylim=c(0,.4), lwd=2,
      ylab="PDF", xlab="x", col="blue", main=hdr)
 lines(k+.04, pdf.h, type="h", lwd=2, col="brown")
 abline(h=0, col="green2")

