To find out how many different selections of three toys Samanta could make from the nine available toys, we can use combinations. The number of ways to choose \( k \) items from \( n \) items is given by the formula:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
In this case, \( n = 9 \) (the total number of toys) and \( k = 3 \) (the number of toys Samanta is selecting). So we need to calculate \( \binom{9}{3} \).
Using the combination formula:
\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} \]
Calculating the factorials, we get:
- \( 9! = 9 \times 8 \times 7 \times 6! \)
- \( 3! = 3 \times 2 \times 1 = 6 \)
- \( 6! \) cancels out in the numerator and the denominator.
Now substituting back into the equation:
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]
So, there are 84 different selections of three toys that Samanta could make.
Now, let's write out Pascal's Triangle from row 0 to row 9 to verify the result:
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Row 6: 1 6 15 20 15 6 1
Row 7: 1 7 21 35 35 21 7 1
Row 8: 1 8 28 56 70 28 8 1
Row 9: 1 9 36 84 126 36 9 1
From Pascal's Triangle, the value in Row 9 and Column 3 is indeed 84, which corresponds to \( \binom{9}{3} = 84 \).
Thus, the final answer is:
\[ \boxed{84} \]
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