let A be a set of 4 digit nos a1,a2,a3,a4 where a1>a2>a3>a4, then n(A) is equal to

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- Nov 13th 2011, 01:22 AM #1

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- Nov 13th 2011, 02:14 AM #2

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## Re: Permu

This can be solved by using combination. Make this a function where X={a1,a2,a3,a4} and Y={0,1,2,...9}. All you have to do is pick 4 numbers from Y. You dont even need to think about the order because the biggest number automatically becomes a1 and the smallest number becomes a4.

- Nov 13th 2011, 05:00 AM #3

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## Re: Permu

Hello, prasum!

Very sloppy wording . . .

$\displaystyle \text{Let }A\text{ be a set of 4-digit numbers}$ composed of the digits $\displaystyle \{a_1,a_2,a_3,a_4\}$

$\displaystyle \text{ where }a_1>a_2>a_3>a_4,$ and each digit is used once.

$\displaystyle \text{ Find }n(A).$

* .The problem said that the 4-digit numbers: .$\displaystyle a_1, a_2, a_3, a_4.$*are*

. . Therefore:.$\displaystyle n(A) \,=\,4.$ .*(duh!)*

* .If the digits can be repeated:.$\displaystyle n(A) \:=\:4^4 \:=\:256.$

Having said all that:.$\displaystyle n(A) \,=\,4!\,=\,\boxed{24}$

BTW the inequality was given to indicate that the digits are distinct.

If the digits were $\displaystyle \{2,7,7,9\}$ or $\displaystyle \{3,3,3,8\}$, the answers would "depend".

- Nov 13th 2011, 05:30 AM #4
## Re: Permu

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