Thread: [SOLVED] permutations and combinatorics

Permutations are formed using all the digits 1,2,3...,9 without repetition.
Determine the number of permutations if:
a) the digits 1,2,3 are together but not necessarily in their natural order
b) even and odd digits alternate
c) the first is an odd digit and the last is on of 1,2,3,4

My thinking for

a) 1,2,3 are together which is 3! and the rest is 6 numbers which is 6! so 3!*6!=4320

Originally Posted by william

Permutations are formed using all the digits 1,2,3...,9 without repetition.
Determine the number of permutations if:
a) the digits 1,2,3 are together but not necessarily in their natural order
b) even and odd digits alternate
c) the first is an odd digit and the last is on of 1,2,

a) 1,2,3 are together which is 3! and the rest is 6 numbers which is 6! so 3!*6!=4320

packets is all - not 6.

Hello, William@

Permutations are formed using all the digits 1,2,3...,9 without repetition.
Determine the number of permutations if:

a) the digits 1,2,3 are together but not necessarily in their natural order

b) even and odd digits alternate

c) the first is an odd digit and the last is one of 1,2,3,4

My thinking for

a) 1,2,3 are together which is 3! and the rest is 6 numbers which is 6!, so 3!*6!=4320
. . . . . Plato pointed out your error.

b) Even and odd digits alternate.

There are 4 even digits and 5 odd digits.
There is one arrangement: .

The odd digits can be scrambled in ways.
The even digits can be scrambled in ways.

Therefore, there are: . such numbers.



c) The first is odd and the last is one of {1,2,3,4}.
There are two cases to consider . . .

(1) The first digit is 1 or 3: 2 choices.
Then the last digit has 3 choices.
And the other seven digits can be arranged in ways.
. . There are: . numbers.

The first digit is 5, 7, or 9: 3 choices.
The last digit has 4 choices.
The other seven digits can be arranged in ways.
. . There are: . numbers.

Therefore, there are: . such numbers.