Answer:
15 subsets of cardinality 4 contain at least one odd number.
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 =
![^3C_1* ^3C_3+^3C_2* ^3C_2+^3C_3* ^3C_1](https://img.qammunity.org/2020/formulas/mathematics/college/flm9f4br6n8dda1n4vaas3zpirdyvhzna3.png)
By using
,
![=3* 1+3* 3+1* 3](https://img.qammunity.org/2020/formulas/mathematics/college/b2ut7s99syu97st05t8ll8dakkef7ae1n6.png)
![=3+9+3](https://img.qammunity.org/2020/formulas/mathematics/college/4j2ygz4wr3n86k0nv670mdxkohmvpz61tm.png)
![=15](https://img.qammunity.org/2020/formulas/mathematics/middle-school/7c0mpo21am95vrizqjloydznkfp76xf6x1.png)
Hence, 15 subsets of cardinality 4 contain at least one odd number.