The number of ways to arrange \( k = k_1 + k_2 + \ldots+ k_m \) items where \( k_1 \) are of one type, \( k_2 \) of another type etc is \( \dfrac{k!}{k_1!k_2! \dots k_m!} \). Now this is obviously an integer as it denotes the number of ways to arrange objects.

Putting \( k = n!, k_1 = k_2 = \dots = k_{(n-1)!} = n \), we see that it satisfies \( k = k_1 + k_2 + \ldots+ k_m \) and number of arrangements is \( \dfrac{(n!)!}{n!n! \dots n!} = \dfrac{(n!)!}{n!^{(n-1)!}} \). Since this is an integer, this implies \((n!)^{(n-1)!} \) divides \( (n!)! \)

EDIT: Changed all the \( (n-1) \) to \( (n-1)! \)
–
Siddhartha Srivastava
·
2 years, 6 months ago

Log in to reply

@Siddhartha Srivastava
–
How is k{1} +k{2} +..... k_{n-1}= k ,i.e, n as it must be equal to n(n-1).
Also question for (n!)! / n!^{(n-1)!} not for (n!)! / n!^{(n-1)}
–
Parv Mor
·
2 years, 6 months ago

Log in to reply

@Parv Mor
–
Yeah. Sorry. I'd put \( (n-1) \) instead of \( (n-1)! \). Sorry again. -.-
–
Siddhartha Srivastava
·
2 years, 6 months ago

## Comments

Sort by:

TopNewestThe number of ways to arrange \( k = k_1 + k_2 + \ldots+ k_m \) items where \( k_1 \) are of one type, \( k_2 \) of another type etc is \( \dfrac{k!}{k_1!k_2! \dots k_m!} \). Now this is obviously an integer as it denotes the number of ways to arrange objects.

Putting \( k = n!, k_1 = k_2 = \dots = k_{(n-1)!} = n \), we see that it satisfies \( k = k_1 + k_2 + \ldots+ k_m \) and number of arrangements is \( \dfrac{(n!)!}{n!n! \dots n!} = \dfrac{(n!)!}{n!^{(n-1)!}} \). Since this is an integer, this implies \((n!)^{(n-1)!} \) divides \( (n!)! \)

EDIT: Changed all the \( (n-1) \) to \( (n-1)! \) – Siddhartha Srivastava · 2 years, 6 months ago

Log in to reply

{1} +k{2} +..... k_{n-1}= k ,i.e, n as it must be equal to n(n-1). Also question for (n!)! / n!^{(n-1)!} not for (n!)! / n!^{(n-1)} – Parv Mor · 2 years, 6 months agoLog in to reply

– Siddhartha Srivastava · 2 years, 6 months ago

Yeah. Sorry. I'd put \( (n-1) \) instead of \( (n-1)! \). Sorry again. -.-Log in to reply