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smc

prealgebra senior

Problem

What is the product of all positive odd integers less than

10000

?

(A)

\frac{10000 !}{( 5000 ! ) ^{2}}

(B)

\frac{10000 !}{2 ^{5000}}

(C)

\frac{9999 !}{2 ^{5000}}

(D)

\frac{10000 !}{2 ^{5000} \cdot 5000 !}

Solution

1 \cdot 3 \cdot 5 \dots 9999 = \frac{1 \cdot 2 \cdot 3 \cdot 4 \dots 10000}{2 \cdot 4 \cdot 6 \dots 10000} = \frac{10000 !}{2 ^{5000} \cdot 1 \cdot 2 \cdot 3 \dots 5000} = \frac{10000 !}{2 ^{5000} \cdot 5000 !}

Therefore the answer is

\frac{10000 !}{2 ^{5000} \cdot 5000 !}

.

Final answer

D