Re: A83: Tutorials...
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Re: A83: Tutorials...
Isn't it 360 because 6!/2!=360 ?
-----Original Message-----
From: Bryan Rabeler <brabeler@isd.ingham.k12.mi.us>
To: assembly-83@lists.ticalc.org <assembly-83@lists.ticalc.org>
Date: Tuesday, June 02, 1998 6:40 PM
Subject: Re: A83: Tutorials...
>
>Icemn711@aol.com wrote:
>
>> That is the right way.
>>
>> > don't you pay too much attention to those 1s now? First you say that
>> > only 3 digits can be the rightmost digit. Then when you've calculated
the
>> > number of possible combinations, you divide that number by two. But by
then
>> > you've already reduced the number of possible combinations once! I'd
say 4
>> > digits can be in the units place (3, 1, 5 and 1), then calculate
>> 5*4*3*2*1*
>> > 4
>> > = 480. _This_ number should be halved -> 240 possible combinations.
>> >
>> > Am I completely wrong?
>> >
>> > Linus
>
>112345 112435 112453 112543 113245 113425 114235
114253
>114325 114523 115243 115423 121345 121435 121453
121543
>123145 123415 123451 123541 124135 124153 124315
124351
>124513 124531 125143 125341 125413 125431 131245
131425
>132145 132415 132451 132541 134125 134215 134251
134521
>135241 135421 141235 141253 141325 141523 142135
142153
>142315 142351 142513 142531 151243 151423 152143
152341
>152413 152431 153241 153421 154123 154213 154231
154321
>
>Ok, the above are all the posibilites for having a 1 in the first place
value.
>(tell me if I missed any) There are 64 total odd numbers. For each
number, you
>could swap the 1 in front with any of 4 other different numbers, so there
should
>be 64*5 = 320 different numbers.
>
>--
>Bryan Rabeler <brabeler@ticalc.org>
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>
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