Find the number of 3-digit numbers, where any two adjacent digits differ by 1.
There are 214 such numbers:
100 101 102 103 104 105 106 107 108 109 110 112 120 121 122 123 124 125 126 127 128 129 132 134 143 145 154 156 165 167 201 210 211 212 213 214 215 216 217 218 219 221 223 230 231 232 233 234 235 236 237 238 239 243 245 254 256 265 267 301 310 312 320 321 322 323 324 325 326 327 328 329 332 334 340 341 342 343 344 345 346 347 348 349 354 356 365 367 401 410 412 421 423 430 431 432 433 434 435 436 437 438 439 443 445 450 451 452 453 454 455 456 457 458 459 465 467 501 510 512 521 523 532 534 540 541 542 543 544 545 546 547 548 549 554 556 560 561 562 563 564 565 566 567 568 569 601 610 612 621 623 632 634 643 645 650 651 652 653 654 655 656 657 658 659 665 667 701 710 712 721 723 732 734 743 745 754 756 760 761 762 763 764 765 766 767 768 769 801 810 812 821 823 832 834 843 845 854 856 865 867 901 910 912 921 923 932 934 943 945 954 956 965 967 Total = 214
Guest I think you may have misread the question.
(unless it was me of course)
Find the number of 3-digit numbers, where any two adjacent digits differ by 1.
Let the first digit by 2,3,4,5,6, or 7
Then there will be 6*2*2=24 such numbers
Say the first digit is 1 or 8
101
121
123
898
878
876
there are 6 of those
Say the first digit is 9
987
989
there are 2 of those
So that is 24+6+2=32