Answer with Step-by-step explanation:
We are given that a set S={a,b,c,d,e}
a.We have n=5
We have to find the number of 2-permutations of S in an organised manner
Using permutation formula
[tex]nP_r=\frac{n!}{r!(n-r)!}[/tex]
r=2
[tex]5P_2=\frac{5!}{3!}[/tex]
[tex]5P_2=\frac{5\times 4\times3!}{3!}[/tex]
[tex]5P_2=20[/tex]
Hence, the total number 2- permutations of S =20
{a,b},{b,c},{c,d},{d,e},{a,c},{a,d},{a,e},{b,d},{b,e},{c,e},{b,a},{c,b},{d,c},{e,d},{c,a},{d,a},{e,a},{d,b},{e,b},{e,c}.
b.We have to find the 2- combinations of S in an organised manner
By using combination formula
[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]
n=5,r=2
[tex]5C_2=\frac{5!}{2!(5-2)!}[/tex]
[tex]5C_2=\frac{5\times 4\times3!}{2!3!}[/tex]
[tex]5C_2=10[/tex]
Hence, the number of 2 -combinations of S =10
{a,b},{b,c},{c,d},{d,e},{a,c},{a,d},{a,e},{b,d},{b,e},{c,e}.
c.We have n=5 r=3
Using combination formula
[tex]5C_3=\frac{5!}{3!(5-3)!}[/tex]
[tex]5C_3=\frac{5\times 4\times 3!}{3!2!}[/tex]
[tex]5C_3=10[/tex]
Hence, the total number of 3- combinations of S =10
{a,b,c},{b,c,d},{c,d,e},{a,b,d},{a,b,e},{b,c,e},{a,c,d},{a,d,e},{b,d,e},{a,c,e}.
d.Number of 2 -combinations of S=Number of 3- combinations of S=10
Combinations is a selection of r elements out of n elements.
When we select 3 elements out of 5 then we get total number of combinations of S=10 and when we select 2 elements out of 5 then we get total number of combinations of S=10
By combinations formula
[tex]5C_3=10[/tex]
[tex]5C_2=10[/tex]
