Permutations :S (1 Viewer)

[Smilebuffalo]

Hey guys i can't do these permutation questions :S Could someone please show me how they are solved.

These questions are all from fitzpatrick 3unit text book exercise 28a)

4. How many five digit numbers can be formed from the digits 2,3,5,6,8,9 if no digit can be used more than once in a number? How many even numbers can be formed? (answer for the 2nd part = 360)

14.b) How many even numbers of 4 digits can be formed with the figures 3,4,7,8 if repetitions are allowed? (answer = 128)

19ii) How many numbers of 7 digits can be formed from the digits 1,2,3,4,5,6,7 if there are not more than 2 digits between 1 and 2? (answer = 3600)

 

[hermand]

4. a.
4. b. 5x4x3x2x3 = 360
you start with the last three, because there are three even numbers, and an even number must be last for the whole number to be even, then, there are 5 numbers left which must consequently be placed in the remaining positions, with one less number each time because one has been used.

14. b. 4x4x4x2 = 128
as there are four options for the first three positions but only two options for the last - must be an even number.

19. ii. (i'm assuming no repetition here)
okay, if one is in the first position, there are three places two can be,
if one is in the second position, there are four places two can me,
third position;
fourth position;
the fifth, sixth and seventh are all copies of the third, second and first, respectively so we can just multiply them by two.
so end eqn will be,

 

[Smilebuffalo]

thanks!