Answer:
15 subsets of cardinality 4 contain at least one odd number.
Step-by-step explanation:
Here the given set,
S={1,2,3,4,5,6},
Since, a set having cardinality 4 having 4 elements,
The number of odd digits = 3 ( 1, 3, 5 )
And, the number of even digits = 3 ( 2, 4, 6 )
Thus, the total possible arrangement of a set having 4 elements out of which atleast one odd number = [tex]^3C_1\times ^3C_3+^3C_2\times ^3C_2+^3C_3\times ^3C_1[/tex]
By using [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex],
[tex]=3\times 1+3\times 3+1\times 3[/tex]
[tex]=3+9+3[/tex]
[tex]=15[/tex]
Hence, 15 subsets of cardinality 4 contain at least one odd number.
