A or B?

A = 64 ! 60 B = 65 ! 64 × 61 A = \frac{64!}{60} \qquad B = \frac{65!}{64\times 61} We can say....

A > B A > B A = B A=B B > A B > A

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4 solutions

A B = 64 ! 60 × 64 × 61 65 ! = 64 ! 65 × 64 ! × 64 × 61 60 = 64 × 61 65 × 60 = 64 × ( 60 + 1 ) ( 64 + 1 ) × 60 = 64 × 60 + 64 64 × 60 + 60 > 1 A > B \dfrac{A}{B} = \dfrac{64!}{60} \times \dfrac{64 \times 61}{65!} = \dfrac{64!}{65 \times 64!} \times \dfrac{64 \times 61}{60} = \dfrac{64 \times 61}{65 \times 60} = \dfrac{64 \times (60 + 1)}{(64 + 1) \times 60} = \dfrac{64 \times 60 + 64}{64 \times 60 + 60} > 1 \Rightarrow \boxed{A > B}

Roger Erisman
Sep 25, 2019

Factor out 59! from both and cancel denominators.

A = 64x63x62x61 = 15249024

B = 65x63x62x60 = 15233400

A>B

Schittering 06
Sep 25, 2019

65! is 65 times bigger than 64! and 64x61=3904, 3904:60 is slightly bigger than 65

Arifin Ikram
Sep 24, 2019

We can write 64 ! 60 \frac{64!}{60} as 65 ! 65 × 60 \frac{65!}{65×60} .

And, 65 ! 65 × 60 \frac{65!}{65×60} > 65 ! 61 × 64 \frac{65!}{61×64}

\Rightarrow 65 ! 3900 \frac{65!}{3900} > 65 ! 3904 \frac{65!}{3904}

So, A > B \boxed{A>B}

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