Make an organized list of the numbers of whites she can choose for each possible number of reds.
Since she must choose at least one of each color, and since the total number has to be less than 100, and since the number of whites has to be less than 10 times the number of reds, the number of reds can be any number from 1 to 98.
For numbers of reds up to 9, the possible numbers of whites are limited by the requirement that the number of whites be less than 10 times the number of reds:
number of reds number of whites number of choices
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1 less than 10; so 1-9 9
2 less than 20; so 1-19 19
3 less than 30; so 1-29 29
... ...
8 less than 80; so 1-79 79
9 less than 90; so 1-89 89
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total ___
For numbers of reds 10 or greater, the requirement that the number of whites be less than 10 times the number of reds is automatically satisfied, because the total number has to be less than 100; the possible numbers of whites are now limited by the requirement that the total number of peppermints be less than 100.
number of reds number of whites number of choices
---------------------------------------------------------
10 less than 90; so 1-89 89
11 less than 89; so 1-88 88
12 less than 88; so 1-87 87
... ...
96 less than 4; so 1-3 3
97 less than 3; so 1-2 2
98 less than 2; so 1 1
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total ____
You can fill in the missing numbers in each table and thus find the answer to the problem.
The sum in each of the two tables is the sum of an arithmetic series, so calculating the sum for each table should be relatively easy.
And the total number of ways is indeed 4446.
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There are many forms of the formula for the sum of the terms of an arithmetic sequence. But rather than a formula in algebraic notation, use a common sense "formula":
The sum of the terms of any arithmetic sequence is
(number of terms) * (average of the first and last terms)
In this problem, there are two arithmetic sequences. Plug in the numbers for each.
In the first, from the analysis for numbers of reds from 1 to 9, there are 9 terms, with first term 9 and last term 89.
In the second, there are 89 terms, with first term 89 and last term 1.