Question

How many ways can a person toss a coin 14 times so that the number of heads is between 6 and 9 inclusive?

Answer 1

How many ways can a person toss a coin 14 times so that the number of heads is between 6 and 9 inclusive?​

the formula of combination is equal to

$\text{nCr}=\frac{n!}{r!(n-r)!}$

For r between 6 and 9

For r=6

n=14

substitute

$14\text{C6}=\frac{14!}{6!(14-6)!}=\frac{14!}{6!(8)!}=\frac{14\cdot13\cdot12\cdot11\cdot10\cdot9}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

14C6=3,003

For r=7

n=14

substitute

$14\text{C7}=\frac{14!}{7!(14-7)!}=\frac{14!}{7!(7)!}=\frac{14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$

14C7=3,432

For r=8

n=14

substitute

$14\text{C8}=\frac{14!}{8!(14-8)!}=\frac{14!}{8!(6)!}=\frac{14\cdot13\cdot12\cdot11\cdot10\cdot9}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

14C8=3,003

For r=9

n=14

substitute

$14\text{C9}=\frac{14!}{9!(14-9)!}=\frac{14!}{9!(5)!}=\frac{14\cdot13\cdot12\cdot11\cdot10}{5\cdot4\cdot3\cdot2\cdot1}$

14C9=2,002

adds the combinations

3,003+3,432+3,003+2,002=11,440

11,440 ways