Grade 12 Data Management Question

Question

Permutations are formed using all the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition. Determine the number of possible permutations in each of the following cases.

a) Even and odd digits alternate. (i.e. 123456789 or 327694185)

b) The digits 123 are together, but not necessarily in that order. (i.e. 123456789 or 481329756)

c) 1 if before 9 but not necessarily adjacent to it (i.e. 192345678 or 234156978)

d) 123 are never all together in any order (i.e. 124356789 or 163924578)

Answer 1

a) Four even digits 2, 4, 6, 8 have 4! permutations. Five odd digits 1, 3, 5, 7, 9 have 5! permutations. Combination of any odd and any even digits permutation will give a unique (=non repeated) number defined by conditions in statement a). Hence, there are 4!5! = 2880 such numbers.

b) There are 6! numbers that consist of six digits 4, 5, 6, 7, 8 and 9. There are 3! permutations (different numbers) from digits 1, 2 and 3. Moreover, each number from these three digits can be positioned before 1st or 2nd or 6th digit or after 6th digit of the number with 6 digits, thus in any of 7 positions. Hence, in case b) there are 7x6!x3! = 3,628,800 numbers.

c) First, we consider the number of different numbers from seven digits 2, 3, 4, 5, 6, 7 and 8. Each permutation is a new number and thus there are 7! numbers.Now we consider position of 9 and number of position of 1 to left from 9. We make a table

Position Number Number of possibilities

of 9 of positions for 1 to insert 9 and 1 before 9

1st 1 1

2nd 2 2

3rd 3 3

... ... ...

8th 8 8

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Total number of the possibilities 36

Hence, there are 36x7! = 181,440 numbers.