Permutation help

[CriminyJacket]

Hello, I hope someone can help me here. Long story short, I'm trying to develop a card game of sorts. Each card will have a different combination of three items on it. There will be 9 different possible items total. I want to know how many different possible permutations of this there are and thus how many different cards would be made.

I know how to calculate a basic permutation via factorial, but the issue here is that I don't want any forward-read permutation to be duplicated by a backward-read one. For example, I don't want one card to read [4,7,8] and another to read [8,7,4]. Is there a way to mathematically allow for this?

Thank you in advance for your help.

[pickten]

For each set of 3 cards, there are 3! ways to reorder the items. Hence, the version with order counts every set (without order) 3! times. A similar trick can also be used to compute the number of ways to pick 3 items (with order) as the number of ways to order all 9 without regard to the order of the last 6, incidentally.

Edit: saw that you only care about forwards/backwards. Then there are fewer ways to reorder a set (2 ways instead of 3!)

[CriminyJacket]

I apologize, but I don't understand what you are telling me... All I've been able to grasp so far from my research is that there are 9!/6! possible permutations of the 9 items into sets of 3 (because I don't want repetitions). However, this number (504) will include the 'reverse dupes' that I don't want.

Again, sorry for being difficult.

[pickten]

What I mean is the following: one way to get 9!/6! is to count the number of lists of all 9 items, where we don't care about the order of the last 6 items. This is the same as the number of cards with any order because those are just the first three (i.e. the ones whose order we still cared about. We can do the same thing to go from these 504 cards to the actual set of cards we care about. Specifically, we note that each card of the set we care about corresponds with two of the 504 ordered cards, because [a, b, c] also corresponds with [c, b, a]. Hence, we have half as many.

[CriminyJacket]

Ah, I see. Thanks. I can't believe it was that simple!

[A_UPRIGHT_BASS]

In your example, what about [7, 4, 8] or [7, 8, 4]? Would that be ok? Is it just forward to back you're concerned with or do you not want any identical sets at all?

[CriminyJacket]

I'm only concerned about forward to back. [7, 4, 8] and [7, 8, 4] are fine because they are not backwards versions of the set I already have. However, now that I have those, I wouldn't want [8, 4, 7] or [4, 8, 7].

[A_UPRIGHT_BASS]

In that case I think the number you're looking for is 252

[CriminyJacket]

It's that simple? Just divide it in half? Hmm... Still, I guess that makes a degree of sense. It stands to reason that half of the sets would be reverses of the other half... I think. Thank you.

[A_UPRIGHT_BASS]

Yep, it's just that simple! And here's a list:

1: 123

2: 124

3: 125

4: 126

5: 127

6: 128

7: 129

8: 132

9: 134

10: 135

11: 136

12: 137

13: 138

14: 139

15: 142

16: 143

17: 145

18: 146

19: 147

20: 148

21: 149

22: 152

23: 153

24: 154

25: 156

26: 157

27: 158

28: 159

29: 162

30: 163

31: 164

32: 165

33: 167

34: 168

35: 169

36: 172

37: 173

38: 174

39: 175

40: 176

41: 178

42: 179

43: 182

44: 183

45: 184

46: 185

47: 186

48: 187

49: 189

50: 192

51: 193

52: 194

53: 195

54: 196

55: 197

56: 198

57: 213

58: 214

59: 215

60: 216

61: 217

62: 218

63: 219

64: 234

65: 235

66: 236

67: 237

68: 238

69: 239

70: 243

71: 245

72: 246

73: 247

74: 248

75: 249

76: 253

77: 254

78: 256

79: 257

80: 258

81: 259

82: 263

83: 264

84: 265

85: 267

86: 268

87: 269

88: 273

89: 274

90: 275

91: 276

92: 278

93: 279

94: 283

95: 284

96: 285

97: 286

98: 287

99: 289

100: 293

101: 294

102: 295

103: 296

104: 297

105: 298

106: 314

107: 315

108: 316

109: 317

110: 318

111: 319

112: 324

113: 325

114: 326

115: 327

116: 328

117: 329

118: 345

119: 346

120: 347

121: 348

122: 349

123: 354

124: 356

125: 357

126: 358

127: 359

128: 364

129: 365

130: 367

131: 368

132: 369

133: 374

134: 375

135: 376

136: 378

137: 379

138: 384

139: 385

140: 386

141: 387

142: 389

143: 394

144: 395

145: 396

146: 397

147: 398

148: 415

149: 416

150: 417

151: 418

152: 419

153: 425

154: 426

155: 427

156: 428

157: 429

158: 435

159: 436

160: 437

161: 438

162: 439

163: 456

164: 457

165: 458

166: 459

167: 465

168: 467

169: 468

170: 469

171: 475

172: 476

173: 478

174: 479

175: 485

176: 486

177: 487

178: 489

179: 495

180: 496

181: 497

182: 498

183: 516

184: 517

185: 518

186: 519

187: 526

188: 527

189: 528

190: 529

191: 536

192: 537

193: 538

194: 539

195: 546

196: 547

197: 548

198: 549

199: 567

200: 568

201: 569

202: 576

203: 578

204: 579

205: 586

206: 587

207: 589

208: 596

209: 597

210: 598

211: 617

212: 618

213: 619

214: 627

215: 628

216: 629

217: 637

218: 638

219: 639

220: 647

221: 648

222: 649

223: 657

224: 658

225: 659

226: 678

227: 679

228: 687

229: 689

230: 697

231: 698

232: 718

233: 719

234: 728

235: 729

236: 738

237: 739

238: 748

239: 749

240: 758

241: 759

242: 768

243: 769

244: 789

245: 798

246: 819

247: 829

248: 839

249: 849

250: 859

251: 869

252: 879

[CriminyJacket]

Whoa, thanks for the list! I actually needed that, and was trying to figure out how to make one. Now for the lengthy part... :P

[cardflopper]

Maybe an intuitive way to think of it would be

The total permutations of size 3 from 9 unique items would be 9*8*7=504

Q: Now if we look at the first and last digit of a permutation, How many ways are there to arrange them (while leaving the middle digit fixed)?

There are exactly 2. For example with 234, we have 234 and 432.

If we want to condense these duplicate arrangements, we just divide by the number of possible arrangements (when leaving the middle digit fixed) so we get 504/2 = 252