Respuesta :
There are 15 writing implements: 8 pencils, 5 ball point pens & 2 felt-tip pens.
(1). Number of ways student can select 6 writing implements out of the total 15 implements is given as;
[tex]C^{15}_{6} = \frac{15!}{6!(15-6)!} =\frac{15!}{6!\times9!}[/tex] [tex]\\ =\frac{15\times 14\times 13\times 12\times 11\times 10\times \not9!}{6\times 5\times 4\times 3\times 2\times \not9!}\\ =7\times 13\times 11\times 5=5005[/tex]
(2). To find the number of ways selection can be made such that no more than one ball point pen is selected, we will add the number of ways in the given two scenarios:
a. When no ball point pen is selected
b. When only one ball point pen is selected
Now, in case a) When no ball point pen is selected than selection can be made by selecting 6 implements out of the 10 implements ( 8 pencils and 2 felt-tip pens) only.
Thus total number of ways in this scenario = [tex]C^{10}_{6} = \frac{10!}{6!(10-6)!} =\frac{10!}{6!\times4!}=\frac{10\times 9\times 8\times 7\times\not6!}{\not6!\times 4\times 3\times 2\times 1}\\ =10\times 3\times 7 = 210 \text{ways}[/tex]
In case b) the selection is made such that there is only one ball point pen. There are 5 ways to select one ball point pen out of 5 ball point pens. And for each of these 5 ways, there will be[tex] C^{10}_{5}[/tex] ways to select remaining 5 writing implements out of the remaining 10 writing implements.
Thus, total number of ways in this scenario = [tex]5\times C^{10}_{5}[/tex] ways
[tex]\\ =5\times \frac{10\times 9\times 8\times 7\times\times 6\times \not5!}{\not5!\times5\times 4\times 3\times 2\times 1}\\ \\=5\times 3\times 2\times 7\times 6 = 1260 \text{ways}[/tex]
Thus, the number of ways selection can be made such that no more than one ball point pen is selected = 210+1260 = 1470 ways.
