albino = amel, they are one in the same.
albino = the inability to produce the pigment melanin.
amel, short for amelanistic = the prefix "a" which means "to lack" and melanistic which means, well melanin.
charcoal = anery type B, they too are one in the same. Charcoal is the "trade name" for anery type B.
Assuming NO known hets for either animal, you are dealing with a trihybrid cross, or in other words 3 separate genetic mutations.
They are: 1 albino, 2, stripe, and 3 anery type B
In breeding the albino stripe to a charcoal you will ONLY produce normal colored and blotched offspring. So in this case, the clutch will come out all 1 morph.
Now, if your albino stripe happens to be heterozygous (aka het) for charcoal, things begin to change some.
Assuming for the following:
albino stripe, het charcoal X charcoal
The resulting clutch will roughly contain phenotypically (phenotypic = visually expressed genes):
75% normal
25% charcoal
Genotypically (Genotypic = actual genetic probabilities)
25% normal, het albino stripe
50% normal, het blizzard stripe (blizzard is the term for an albino charcoal)
25% charcoal, het albino stripe
I can go on and on with this...
As daddyo (sic) pointed out, check out the corn calculator he linked you to. You can adjust for potential hets to see what the possible outcomes are.
THE BIGGEST TAKE HOME HERE THOUGH IS THAT THIS IS ALL PROBABILITY/STATISTICS. Even if your albino stripe was het charcoal and your charcoal was het albino stripe, it wouldn't necessarily mean that you'd produce a blizzard or even a blizzard stripe. It's a flip of the coin with each egg. The percentages that the corn calc will show you is the probability for each individual egg and not the whole clutch. Over the course of producing 1000 or even 1,000,000 eggs, then and only then will the percentages start to be a reflection of the clutch as a whole. This is only because your sampling size is large and the Law of Averages will start to factor in.