A student has a cup with 14 writing implements: 7 pencils, 5 ball point pens, and 2 felt-tip pens.
In how many ways can the student select 4 writing implements? = C(14,4) = 1001
In how many ways can the selection be made if no more than one ball point pen is selected?
To find the number of ways the selection can be made if no more than one ball point pen is selected, we need to consider two scenarios:
Scenario 1: No ball point pen is selected
Scenario 2: Exactly one ball point pen is selected
For Scenario 1, we can choose 4 writing implements from the remaining 9 options (7 pencils and 2 felt-tip pens), which can be done in C(9,4) ways.
For Scenario 2, we need to choose 3 more writing implements from the remaining 8 options (5 ball point pens, 7 pencils, and 2 felt-tip pens), which can be done in C(8,3) ways. We also have to multiply this by the number of ways to select the one ball point pen, which is 5 (since there are 5 ball point pens to choose from).
Therefore, the total number of ways to select 4 writing implements with no more than one ball point pen is:
C(9,4) + 5 * C(8,3) = 126 + 5 * 56 = 126 + 280 = 406 ways.
To find the number of ways the selection can be made if no more than one ball point pen is selected, we need to consider two cases:
Case 1: No ball point pen is selected
In this case, the student can choose 4 writing implements from the remaining 12 (7 pencils and 2 felt-tip pens). The number of ways to choose 4 from 12 is given by C(12,4).
Case 2: Exactly one ball point pen is selected
In this case, the student can choose 3 more writing implements from the remaining 11 (7 pencils, 5-1=4 ball point pens, and 2 felt-tip pens). The number of ways to choose 3 from 11 is given by C(11,3).
To find the total number of ways, we sum up the number of ways from both cases:
Total number of ways = C(12,4) + C(11,3)
Let's calculate it step by step:
1. Calculate C(12,4):
C(12,4) = 12! / (4!(12-4)!)
= 12! / (4!8!)
= (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1)
= 495
2. Calculate C(11,3):
C(11,3) = 11! / (3!(11-3)!)
= 11! / (3!8!)
= (11 * 10 * 9) / (3 * 2 * 1)
= 165
3. Add the two values together:
Total number of ways = 495 + 165
= 660
Therefore, there are 660 ways to make the selection if no more than one ball point pen is selected.
First of all, C(14,4) is not 1001.
The only combinations possibilties are:
PPP (3 pencils)
PPF (2 pencils and 1 felt-tip)
PPB (2 pencils and 1 ballpoint, etc)
BBB
BBP
BBF
FFP
FFB (the least likely result)
PBF
Different permutations of the same group of three don't count as being different.
That is a total of only 9 possibilities.
The probabilities of each will not be the same.