$$ 1 \leftarrow $$ How many distinguishable DNA sequences can be formed using three As, two Cs, two Gs, and three Ts? There are $$ \square $$ possible distinguishable sequences that can be formed. (Type a whole number.)

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To find the number of distinguishable DNA sequences that can be formed using three As, two Cs, two Gs, and three Ts, we need to calculate the total number of permutations of these letters.

The total number of letters is $$3 + 2 + 2 + 3 = 10$$.

The number of distinguishable DNA sequences can be calculated using the formula for permutations:

$$ \text{Number of permutations} = \frac{10!}{3! \cdot 2! \cdot 2! \cdot 3!} $$

Where:
- $$10!$$ represents the total number of permutations of all 10 letters.
- $$3!$$ represents the number of permutations of the three As.
- $$2!$$ represents the number of permutations of the two Cs and the two Gs.

Let's calculate this:
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{10!}{\left(3!\times 2!\times 2!\times 3!\right)}$$
- step1: Remove the parentheses:
  $$\frac{10!}{3!\times 2!\times 2!\times 3!}$$
- step2: Multiply the terms:
  $$\frac{10!}{144}$$
- step3: Calculate the value:
  $$\frac{3628800}{144}$$
- step4: Reduce the fraction:
  $$25200$$
The number of distinguishable DNA sequences that can be formed using three As, two Cs, two Gs, and three Ts is 25,200.
There are 25,200 distinguishable DNA sequences that can be formed.

How many distinguishable DNA sequences can be formed using three As, two Cs, two Gs, and three Ts?
There are possible distinguishable sequences that can be formed.
(Type a whole number.)

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How many distinguishable DNA sequences can be formed using three As, two Cs, two Gs, and three Ts? There are possible distinguishable sequences that can be formed. (Type a whole number.)