Question

The list of sequences of Heads (H) and Tails (T) in four coin flips with various numbers of heads is as follows: 0 Heads TTTT 1 Head HTTT THTT TTHT TTTH 2 Heads HHTT HTHT THHT HTTH THTH TTHH 3 Heads HHHT HHTH HTHH THHH 4 Heads HHHH Write out the list of sequences of Heads (H) and Tails (T) in five coin flips with 0 Heads, 1 Head, 2 Heads, 3 Heads, 4 Heads and 5 Heads. Describe how we can use the information about four coin flips to list the sequences for five coin flips. Describe why the number of sequences of Heads and Tails on n coin flips that have k heads is the same as the number of sequences of Heads and Tails on n − 1 coin flips that have k heads plus the number of sequences of Heads and Tails on n − 1 coin flips that have k − 1 head

Answer 1

Each of the 4 coin cases will have two sub cases each corresponding to H and a T. Hence, using the 4 coins 16 cases, we end up 32 cases with 5 coins.

0H: TTTTT

1H: HTTTT, THTTT, TTHTT,TTTHT, TTTTH

2H: HHTTT, HTHTT, HTTHT, HTTTH, THHTT, THTHT, THTTH, TTHHT, TTHTH, TTTHH

3H: TTHHH, THTHH, THHTH, THHHT, HTTHH, HTHTH, HTHHT, HHTTH, HHTHT, HHHTT

4H: THHHH, HTHHH, HHTHH, HHHTH, HHHHT,   

5H: HHHHH

b)Number of sequences on n coin flips that have k heads=nCk

number of sequences on (n-1) flips that have k heads=(n-1)Ck

number of sequences on (n-1) flips that have (k-1) heads=(n-1)C(k-1)