As I was

Probability Level 1

As I was going to St. Ives,

I met a man with seven wives.

I asked and he said "It's more fun

Than to marry only one."

Assuming he didn't have a marriage where he wed two or more of the wives simultaneously, in how many different orders could he have married his seven wives?


The answer is 5040.

Denton Young by Denton Young , 47, USA

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2 solutions

Chew-Seong Cheong
Jun 20, 2016

O M G ! T h a t ′ s e x a c t l y 7 ! E q u a l s 5040 , n o m o r e a n d n o l e s s ! \mathscr {OMG}! \ \mathscr That's \ exactly \ 7! \\ \mathscr Equals \ \boxed{5040}, \ no \ more \ and \ no \ less!

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Denton Young
Jun 19, 2016

He can choose the first wife to marry in 7 different ways. Then he has six different possibilities for wife #2 (etc.)

The answer is 7! = 5040 different orderings.

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