Answer: (a) 5 (b)1 (c)5 (d)120
Step-by-step explanation:
Formula for combinations: [tex]C(n,r)=\dfrac{n!}{r!(n-r)!}[/tex]
Formula for permutations: [tex]C(n,r)=\dfrac{n!}{(n-r)!}[/tex]
(a) C(5, 4) = [tex]\dfrac{5!}{4!(5-4)!}=\dfrac{5\times4!}{4! (1)}=5[/tex]
(b) C(5, 0)
= [tex]\dfrac{5!}{0!(5-0)!}=\dfrac{5!}{5! }=1[/tex]
(c) P(5, 1) = [tex]\dfrac{5!}{(5-1)!}=\dfrac{5\times4!}{4! }=5[/tex]
(d) P(5, 5) = [tex]\dfrac{5!}{(5-5)!}=\dfrac{5!}{1}=5!= 5\times4\times3\times2\times1=120[/tex] [as 0!=1]
Hence, the required answer is
(a) 5 (b)1 (c)5 (d)120
