+1443 A coin is tossed $12$ times. How many different sequences of tosses could there have been, if the number of heads was at least $5?$
+129850 Number of possible outcomes = 2^12
Number where 0 heads were tossed = 1
Number where 1 head was tossed = 12
Number where 2 heads were tossed = C (12 ,2) = 66
Number where 3 heads are tossed = C (12,3) = 220
Number where 4 heads are tossed= C (12, 4) = 495
Number of different sequences where at least 5 heads were tossed =
2^12 - 1 - 12 - 66 - 220 - 495 = 3302
+57 You could use pascal's triangle and find out what it is, looking at the first or last 6 numbers in the 13th row, including the ones at the top, add them together, and substract that from 2^12
