Determine the following regarding number of ways things can occur. 10.1 Order unimportant and no replacement, select 1 centre, 2 guards, and 2 forwards for a basketball team given three centre players, 5 forward players, and 6 guard players from which to choose. 10.2 Ordee unimportant and no replacement, ensure that at least 1 of the next 3 passengers boarding the aircraft has a window seat given that 8 seats are left with 4 of those being window seats. 10.3 Possible lottery numbers if each lottery ticket purchaser selects a six digit number with digit 0 through 9 allowed 10.4 Number of unique orderings without replacement of nine marbles if they include 4 identical red, 3 identical white, and 2 identical blue. 10.5. Difference in number of selections assuming 5 unique items without replacement as between selecting 5 or 0 at a time.
Determine the following regarding number of ways things can occur. 10.1 Order unimportant and no replacement, select 1 centre, 2 guards, and 2 forwards for a basketball team given three centre players, 5 forward players, and 6 guard players from which to choose. 10.2 Ordee unimportant and no replacement, ensure that at least 1 of the next 3 passengers boarding the aircraft has a window seat given that 8 seats are left with 4 of those being window seats. 10.3 Possible lottery numbers if each lottery ticket purchaser selects a six digit number with digit 0 through 9 allowed 10.4 Number of unique orderings without replacement of nine marbles if they include 4 identical red, 3 identical white, and 2 identical blue. 10.5. Difference in number of selections assuming 5 unique items without replacement as between selecting 5 or 0 at a time.
10.1
As order is unimportant, we do combinations.
For center, the are 3C1 ways = 3 ways.
For guards, there are 6C2 ways = 15 ways.
For forwards, there are 5C2 ways = 10 ways.
Thus, the total number of ways a 5-man lineup can be made is 3*15*10 = 450 ways. [ANSWER, 450]
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10.2
\"At least one has a window seat\" is the complement of \"none of them has a window seat\".
There are 4 non-window seats for these 3 persons.
Thus,there are a total of 4C3 = 4 ways to arrange so that none of them have window seats.
There are 8C3 = 56 ways to arrange these persons.
Thus, the ways in which at least 1 of them has a window seat is 56 - 4 = 52 ways. [ANSWER, 52]
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