# Thread:

1. ## Permutation and Combination Help!

Question:
A coin is tossed 10 times.
(a) How many different sequences of heads and tails are possible?
(b) How many different sequences containing six heads and four tails are possible?
(c) What is the probability of getting si heads and four tails?

Attempt:
(a) = 0 Heads or 1 Heads or 2 Heads or 3 Heads or 4 Heads or 5 Heads or 6 Heads or 7 Heads or 8 Heads or 9 Heads or 10 Heads.
$= 10C0 + 10C1 + 10C2 + 10C3 + 10C4 + 10C5 + 10C6 + 10C7 + 10C8 + 10C9 + 10C10$
$= 1024$

(b) $= 10C6+10C4 = 420$

(c) No Idea!

2. Hello, looi76!

A coin is tossed 10 times.
(a) How many different sequences of heads and tails are possible?
Your answer is correct, but there's an easier approach.

For each coin there are two possible states: Heads or Tails.

. . With ten coins, there are: . $2^{10} \:=\:1024$ possible outcomes.

(b) How many different sequences containing 6 Heads and 4 Tails are possible?
There are: . ${10\choose6,4} \:=\:210$ ways.

(c) What is the probability of getting 6 Heads and 4 Tails?
There are 210 ways to get 6 Heads and 4 Tails
. . out of the 1024 possible outcomes.

Therefore: . $P(\text{6 Heads, 4 Tails}) \;=\;\frac{210}{1024} \;=\;\frac{105}{512}$
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