Help: Perms and combs (1 Viewer)

[braintic]

This can be answered with more conventional Ext 1 logic:

10 + 10P2 + 10C2 + 2(10C3) + 10C4

See if you can work out the logic behind that calculation.

 

[kawaiipotato]

How did you do it with stars and bars? I can't really visualise it

 

[Zlatman]

This can be answered with more conventional Ext 1 logic:

10 + 10P2 + 10C2 + 2(10C3) + 10C4

See if you can work out the logic behind that calculation.

10C4: four distinct numbers (pick 4 different numbers from 10)
3 x (10C3): 3 groups total: one group of two digits and two with one (pick 3 numbers from 10, pick one of the numbers to have two digits)
10C2: 2 groups of two digits (pick 2 numbers from the 10)
10P2: 2 groups: one with 3 digits, one with 1 (pick 2 numbers from 10, order matters because one is the group of three, the other is the group of one)
10: 4 digits of the same number (pick 1 number from 10)

= 715

 

[InteGrand]

How did you do it with stars and bars? I can't really visualise it

Imagine 10 'spaces' corresponding to the digits 0-9, and four dots, which we put into these spaces. The number of dots in each digit's space will be the number of times that digit was selected out of our 4 selected ones.

I.e. • | | | •• | | • | | | | .

The above corresponds to the choice: one 0, two 3's and one 5.

There 9 bars (as there are ten digits) and 4 identical dots, so we just arrange those, as each arrangement of them corresponds to exactly one way of selecting the four numbers (and clearly all the possible ways of selecting them are mapped to exactly one arrangement of the above; in other words there is a bijection between the set of arrangements of the above and the set of possible choices of the digits).