no, not the whole answer but some highlights: There are

choices for the first vector (since zero cannot be part of a basis), then there are

choices for the second vector (all the available vectors minus those that are a scalar multiple of the first one), then there are

choices for the third one (all the possible vectors minus all the scalar multiples of the first two), etc.

All in, there are

...try to end the argument by yourself now.

Tonio

ps ...הרבה יותר כיף לגלות את הפתרון בעצמך, אפילו אם זה קשה, מאשר לקבל את הכל מוכן