Assume that the student has a cup with 14 writing implements: 7 pencils, 5 ball point pens, and 2 felt-tip pens.
(1) In how many ways can the student select 8 writing implements?
(2) In how many ways can the selection be made if no more than one ball point pen is selected?
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To find the number of ways the student can select 8 writing implements, we can use a combination formula. The formula for combinations is given by:
C(n, r) = n! / (r! * (n-r)!)
Where n is the total number of objects (writing implements in this case) and r is the number we want to select.
(1) In this case, the student wants to select 8 writing implements out of a total of 14. Using the combination formula:
C(14, 8) = 14! / (8! * (14-8)!)
Simplifying the equation:
C(14, 8) = 14! / (8! * 6!)
Now, let's calculate this value:
14! = 14 * 13 * 12 * 11 * 10 * 9 * 8!
6! = 6 * 5 * 4 * 3 * 2 * 1
Substituting these values into the equation:
C(14, 8) = (14 * 13 * 12 * 11 * 10 * 9 * 8!) / (8! * (6 * 5 * 4 * 3 * 2 * 1))
Simplifying further:
C(14, 8) = (14 * 13 * 12 * 11 * 10 * 9) / (6 * 5 * 4 * 3 * 2 * 1)
By canceling out common factors:
C(14, 8) = (14 * 13 * 12 * 11 * 10 * 9) / (6 * 5 * 4 * 3 * 2)
Evaluating the expression:
C(14, 8) = 3003
Therefore, the student can select 8 writing implements in 3003 different ways.
(2) If no more than one ball point pen is allowed, we need to consider two cases: selecting 0 ball point pens and selecting 1 ball point pen.
For case 1 (0 ball point pen), we can select 8 writing implements from a total of 14 (excluding the 5 ball point pens). Similar to the previous calculation, using the combination formula:
C(9, 8) = 9! / (8! * (9-8)!) = 9
For case 2 (1 ball point pen), we need to select 7 additional writing implements from the remaining (14 - 1) = 13 total writing implements (excluding the selected ball point pen). Using the combination formula again:
C(13, 7) = 13! / (7! * (13-7)!) = 13 * 12 * 11 * 10 * 9 * 8 / (7 * 6 * 5 * 4 * 3 * 2 * 1) = 1716
Finally, the total number of ways to make the selection is the sum of both cases:
Total ways = Case 1 + Case 2 = 9 + 1716 = 1725
Therefore, if no more than one ball point pen is selected, the student can make the selection in 1725 different ways.