How many 10-digit numbers How many $10$-digit numbers have two digits $1$, two digits $2$, three digits $3$, three digits $4$ so that between the two digits $1$ it has at least other two digits and between two digits $2$ it has at least other two  digits (not necessarily distinct)? Thanks!
 A: I am not sure if I get your problem right, but from my understanding I think a very simple solution exists - number of 10 digits numbers with two 1 digit, two 2 digit, three 3 digit and three 4 digit - 10!/(2! 2! 3! 3!) = 25200, this includes all possible numbers, now we consider the cases where between two digit 1 exactly one other digit is present - to determine this we consider first 1 is at first position and second 1 is at position third (eg 1-1-------) now for 8 remaining position two digit 2, three digit 3 and three digit 4 can be arranged - 8!/(2!3!3!) = 560, so for each of the 8 such combination [(1st - 3rd) (3rd - 5th) (2nd - 4th) ..... ] we get 8*560 = 4480 cases, now if we consider the case where two digit 1 are placed consecutive we get 9 cases and for each case 560 distinct arrangements, so total 4480+9*560=9520 distinct cases are there where between two digit 1 less than 1 other digit is present, subtracting the number we get 25200-9520=15680 cases where between two digit 1 two or more other digit (not necessarily distinct) is present, similarly for two digit 2 we also get same number of cases so total number of cases(25200) - [[number of cases where between two digit 1 one or less digit is present (9520)] or [number of cases where between two digit 2 one or less digit is present (9520)]] = 6160 numbers where between the two digits 1 it has at least other two digits and between two digits 2 it has at least other two digits (not necessarily distinct) 
