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Question Evaluate and simplify the expression 4! 5!3! 7!\frac{4! \, 5!}{3! \, 7!} to 1584\frac{15}{84}.

Studdy Solution

STEP 1

1. The notation n!n! represents the factorial of nn, which is the product of all positive integers from 11 to nn.
2. Factorials can be simplified by canceling common factors in the numerator and denominator.

STEP 2

1. Expand the factorials in the numerator and denominator.
2. Cancel common factors.
3. Simplify the resulting expression.

STEP 3

Expand the factorials in the numerator and denominator.
4!5!3!7!=(4⋅3⋅2⋅1)(5⋅4⋅3⋅2⋅1)(3⋅2⋅1)(7⋅6⋅5⋅4⋅3⋅2⋅1) \frac{4!5!}{3!7!} = \frac{(4 \cdot 3 \cdot 2 \cdot 1)(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{(3 \cdot 2 \cdot 1)(7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}

STEP 4

Notice that 4!4! is a part of both 5!5! and 7!7!, so we can simplify the expression by canceling out 4!4! from both the numerator and the denominator.
4!5!3!7!=4!⋅(5⋅4⋅3⋅2⋅1)3!⋅(7⋅6⋅5⋅4⋅3⋅2⋅1)=57⋅6 \frac{4!5!}{3!7!} = \frac{4! \cdot (5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{3! \cdot (7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)} = \frac{5}{7 \cdot 6}

STEP 5

Cancel common factors between the numerator and the denominator.
57⋅6=542 \frac{5}{7 \cdot 6} = \frac{5}{42}

STEP 6

The expression is already simplified, so this is the final answer.
542 \frac{5}{42}

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