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 November 19th, 2011, 03:26 AM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 permutation $\tex{total no. of five digit no. of different digit in which }$ $\tex{the digit in the middle is largest}$
 November 19th, 2011, 06:55 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: permutation \begin{align*}\sum_{n=1}^i (n\,-\,1)n(n\,-\,2)(n\,-\,3)\,&=\,\sum_{n=1}^i n^4\,-\,6n^3\,+\,11n^2\,-\,6n \\ &=\,\frac{i^5}{5}\,+\,\frac{i^4}{2}\,+\,\frac{i^3} {3}\,-\,\frac{i}{30}\,-\,6$$\frac{i^4}{4}\,+\,\frac{i^3}{2}\,+\,\frac{i^2 }{4}$$\,+\,11$$\frac{i^3}{3}\,+\,\frac{i^2}{2}\,+\ ,\frac{i}{6}$$\,-\,6$$\frac{i(i\,+\,1)}{2}$$ \\ &=\,\frac{i^5}{5}\,-\,i^4\,+\,i^3\,+\,i^2\,-\,\frac{6i}{5} \\ &=\,3024\,\leftarrow\,i\,=\,8\end{align*}
 November 19th, 2011, 07:42 AM #3 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: permutation $\sum_{n=4}^{9}\binom{n}{4}4!=6048$ Assuming first digit can be zero.
 November 19th, 2011, 07:59 AM #4 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: permutation ${10 \choose 5}$ ways to pick 5 different digits out of 10 (0 to 9) They can be arranged in 5! ways. one in five numbers will have the middle digit largest. Would yield ${10 \choose 5} \cdot 5! \cdot \frac{1}{5}= 6048$ However You might exclude numbers with 0 as leading digit. ${9 \choose 4}$ of these with different digits. Leaving only $\left({10 \choose 5} - {9 \choose 4}\right) \cdot 5! \cdot \frac{1}{5}= 3024$
 November 19th, 2011, 09:44 AM #5 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 Re: permutation Well I'm getting a different answer. $\sum_{n=3}^8 n \cdot _nP_3 = 5292$ I'm assuming that the first digit is not zero and that the middle digit (the largest) must be 4,5,6,7,8, or 9.
November 19th, 2011, 10:45 AM   #6
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: permutation

Hello, panky!

I agree with mrtwhs.

Quote:
 $\tex{Total number of five digit numbers of different digit}$ $\text{ in which the middle digit is he largest}$

I will assume that the number must not begin with zero.

There are two cases to consider.

$\text{[1] A zero (0) is included.}$
[color=beige]. . [/color]$\text{There are: }\,{9\choose4} \,=\,126\text{ choices for the other four digits.}$

$\text{The largest is placed in the middle: }\:\_\;\_\;L\;\_\:\_$
[color=beige]. . [/color]$\text{There are }3\text{ choices for the first digit.}$
[color=beige]. . [/color]$\text{The other three digits can be placed in }3!\text{ orders.}$

$\text{There are: }\:126\,\cdot\,3\,\cdot\,3! \:=\:2268 \text{ numbers with a zero.}$

$\text{[2] Zero is not included.}$
[color=beige]. . [/color]$\text{There are: }\,{9\choose5} \,=\,126 \text{ choices for the 5 digits.}$

$\text{The largest is in the middle: }\:\_\;\_\;L\;\_\;\_$
[color=beige]. . [/color]$\text{There are: }4! \text{ ways to place the other 4 digits.}$

$\text{There are: }\:126 \,\cdot\,24 \,=\,3024\text{ numbers without a zero.}$

$\text{Therefore, there are: }\:2268\,+\,3024 \:=\:5292\text{ such numbers.}$

November 19th, 2011, 10:56 AM   #7
Senior Member

Joined: Oct 2011
From: Belgium

Posts: 522
Thanks: 0

Re: permutation

Quote:
 Originally Posted by Hoempa $\left({10 \choose 5} - {9 \choose 4}\right) \cdot 5! \cdot \frac{1}{5}= 3024$
Hoempa,

It should be ${10 \choose 5} \cdot 5! \cdot \frac{1}{5} - {9 \choose 4} \cdot 4! \cdot \frac{1}{4}= 5292$

And then you have the same result as soroban and mrtwhs.

 November 20th, 2011, 03:28 AM #8 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: permutation Thanks, wnvl! So it is like: if 0 is the leading digit, there are 9 choices for the 4 digits, which can be arranged in 4! ways. In 1 in 4 cases, the 2nd of the 4 (middle digit of the no.) will be the largest. ${9 \choose 4} \cdot 4! \cdot \frac{1}{4}$ So ${10 \choose 5} \cdot 5! \cdot \frac{1}{5} - {9 \choose 4} \cdot 4! \cdot \frac{1}{4}= 5292$ possibilities.
 November 21st, 2011, 11:11 PM #9 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Re: permutation Thanks to all for nice explanation. answer given is =5292

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