9514 1404 393
Answer:
210
Step-by-step explanation:
The number of combinations of 10 things taken 6 at a time is ...
10!/(6!(10-6)!) = 210
210 bit strings of length 10 will have 6 1-bits.
Using combination
[tex]\\ \sf\longmapsto {}^{10}C_6[/tex]
We know
[tex]\boxed{\sf {}^nC_r=\dfrac{n!}{r!(n-r)!}}[/tex]
[tex]\\ \sf\longmapsto \dfrac{10!}{6!(10-6)!}[/tex]
[tex]\\ \sf\longmapsto \dfrac{10!}{6!(4!)}[/tex]
[tex]\\ \sf\longmapsto 210[/tex]
