True or False

Algebra Level 3

True or False?

∏ k = 1 n k ! = ∏ k = 1 n k n + 1 − k \large \displaystyle \prod_{k=1}^{n} k!=\displaystyle \prod_{k=1}^{n} k^{n+1-k}

False True

Hjalmar Orellana Soto by Hjalmar Orellana Soto , 24, Chile

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2 solutions

∏ k = 1 n k ! = 1 ! ⋅ 2 ! ⋅ 3 ! ⋯ n ! = 1 ⋅ ( 1 ⋅ 2 ) ⋅ ( 1 ⋅ 2 ⋅ 3 ) ⋯ ( 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋯ n ) \displaystyle \prod_{k=1}^{n} k!=1!\cdot 2!\cdot 3!\cdots n!=1\cdot (1 \cdot 2)\cdot (1\cdot 2\cdot 3)\cdots (1\cdot 2 \cdot 3\cdot 4 \cdots n) = 1 n ⋅ 2 n − 1 ⋅ 3 n − 2 ⋯ ( n − 1 ) 2 ⋅ n 1 = ∏ k = 1 n k n + 1 − k =1^{n}\cdot 2^{n-1} \cdot 3^{n-2} \cdots (n-1)^{2} \cdot n^{1} =\displaystyle \prod_{k=1}^{n} k^{n+1-k}

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Sabhrant Sachan
Jan 5, 2017

1 ! ⋅ 2 ! ⋅ 3 ! ⋯ ( n − 1 ) ! ⋅ n ! = 1! \cdot 2! \cdot 3! \cdots (n-1)! \cdot n! = ( n ⋅ ( n − 1 ) ⋅ ( n − 2 ) ⋯ 3 ⋅ 2 ⋅ 1 ( n − 1 ) ⋅ ( n − 2 ) ⋯ 3 ⋅ 2 ⋅ 1 ( n − 2 ) ⋯ 3 ⋅ 2 ⋅ 1 ⋮ 3 ⋅ 2 ⋅ 1 2 ⋅ 1 1 ) n rows = 1 n ⋅ 2 n − 1 ⋅ 3 n − 2 ⋯ ( n − 1 ) 2 ⋅ n 1 \left( \begin{matrix} \begin{aligned} n \cdot (n-1) \cdot (n-2) \cdots 3 \cdot 2 & \cdot 1 \\ (n-1) \cdot (n-2) \cdots 3 \cdot 2 & \cdot 1 \\ (n-2) \cdots 3 \cdot 2 & \cdot 1 \\ & \vdots \\ 3 \cdot 2 & \cdot 1 \\ 2 & \cdot 1 \\ & 1 \end{aligned} \end{matrix} \right) \hspace{1mm} \small n \text{ rows} \\ = 1^{n}\cdot 2^{n-1}\cdot 3^{n-2} \cdots (n-1)^{2} \cdot n^{1}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution, Sambhrant

Hjalmar Orellana Soto - 4 years, 5 months ago

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Thank you :) , your solution is also good but i think vertical expansion of the product is easy to understand.

Sabhrant Sachan - 4 years, 5 months ago

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