The possible arrangements are of the form 35 _ Can be chosen in 6 ways. 35 _ _ We can choose 2 out of the remaining 6 in 6C2=15ways. We remove 1 case where 7 and 8 are together to get 14 ways. 35 _ _ _We can choose 3 out of the remaining 6 in 6C3= 20ways. We remove 4 cases where 7 and 8 are together to get 16 ways. 35 _ _ _ _We can choose 4 out of the remaining 6 in 6C4=15ways. We remove 6 case where 7 and 8 are together to get 9 ways. 35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways. Thus, total number of cases =6+14+16+9+2=47 Alternatively, The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8. Considering the cases, only 7 is selected. We can select a maximum of 7 digit numbers. We must select 3, 5, and 7. Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have Each of the numbers can either be selected or not selected and we have 4 numbers : Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2×2×2×2=16 possibilities. SImilarly, including only 8, we have 16 more possibilities. Cases including neither 7 nor 8. We must have 3 and 5 in the group but there must be no 7 and 8 in the group. Hence we have 3 5 _ _ _ _. For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1,2,4,6=16possibilities But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number. Hence 16−1=15cases. Total =16+15+16=47possibilities