Question

please help me in these questions please​

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$Q.1 \\ there \: are \: \: 21 \: \: points \: \: on \: a \: certain \\ circle \: \\ so \: we \: know \: that \\ \: a \: chord \: \: can \: be \: drawn \: to \: any \: \: circle \\ through \: two \: distinct \: points \\ lying \: on \: the \: circle \\ \\ so \: thus \: then \\ no \: of \: chords \: that \: can \: be \: drawn \: to \: \\ given \: circle \: can \: be \: done \\ \: as \: follows : - \\ \\ \frac{21 \:! }{(21 - 2)!.2 \:! } \\ \\ = \frac{21 \: \times 20 \times 19 \:! }{19 \: ! \times 2} \\ \\ = \frac{21 \times 20}{2} \\ \\ = 21 \times 10 \\ \\ = 210 \: \: chords$

$Q.2 \\ a \: team \: of \: 3 \: boys \: and \: \: 3 \: girls \\ is \: to \: be \: made \: from \: \\ 5 \: boys \: and \: 4 \: girls \\ \\ therefore \: this \: can \: be \: done \: in \\ following \: ways \\ \\ = \frac{5 \:! }{(5 - 3)! \times 3!} \: \times \frac{4!}{(4 - 3)! \times 3!} \\ \\ = \frac{5 \times 4 \times 3 \times 2!}{2! \times 3 \times 2} \times \frac{4 \times 3!}{1! \times 3!} \\ \\ = (5 \times 2) \times (4) \\ \\ = 10 \times 4 \\ \\ = 40 \: ways$