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If there are $$M$$ ways from $$P$$ to $$Q$$ and $$N$$ ways to go from $$Q$$ to $$R$$, then the total number of ways of going from $$P$$ to $$R$$ is $$M \times N$$
Addition principle: If event A can occur in m ways and event B in n ways and they are mutually exclusive, total = m+n
Six friends Ram, Shyam, Ramesh, Mahesh, Sundar and Raj are seated around a bonfire. As Ram and Shyam are best friends, they always sit together. In how many ways can we arrange the friends so that Ram and Shyam sit together?
Consider Ram and Shyam as one unit and arrange the remaining 4 + 1 combination around a circle in (5-1)! ways.
Ram and Shyam can be switched within the unit in 2 ways.
Therefore, the total number of arrangements is 2*(5-1)! = 48 ways.
The 100m dash at the school sports meet has 10 runners. If the first three finishers win a gold, silver and bronze medal, how many different combinations of medal winners exist
The first place finisher can be selected in 10 ways, second in 9 and third in 8. Thus the number of ways= 10*9*8 =720
A state decides to create license plates with six different alphanumeric characters such that at least 2 characters are alphabets and at least two are digits. How many different license plates can they produce?
As 2 characters are alphabets and 2 are numbers, the remaining 2 can be either numbers or alphabets. Hence, three cases exist.
Case 1: Both the remaining characters are alphabets
Hence the plate will have 4 alphabets and 2 numbers. The 4 alphabets can be chosen in $$^{26}C_{4}$$ ways. The 2 numbers can be chosen in $$^{10}C_{2}$$ ways. Hence, number of ways of selection under case 1 = $$^{26}C_{4}$$ * $$^{10}C_{2}$$
Case 2: Both the remaining characters are numbers
Hence the plate will have 2 alphabets and 4 numbers. The 2 alphabets can be chosen in $$^{26}C_{2}$$ ways. The 4 numbers can be chosen in $$^{10}C_{4}$$ ways. Hence, number of ways of selection under case 2 = $$^{26}C_{2}$$ * $$^{10}C_{4}$$
Case 3: One character is a number and one is an alphabet
Hence the plate will have 3 alphabets and 3 numbers. The 3 alphabets can be chosen in $$^{26}C_{3}$$ ways. The 3 numbers can be chosen in $$^{10}C_{3}$$ ways. Hence, number of ways of selection under case 3 = $$^{26}C_{3}$$ * $$^{10}C_{3}$$
In each of the cases, as there is no repetition, the arrangement can be done in 6! ways.
Hence, total number of plates = 6! * ($$^{26}C_{4}$$ * $$^{10}C_{2}$$ + $$^{26}C_{2}$$ * $$^{10}C_{4}$$ + $$^{26}C_{3}$$ * $$^{10}C_{3}$$)
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