Question

11.3.35

Answer 1

Answer:

The different selections possible are 12,21,759.

Step-by-step explanation:

Given:

Number of different numbers available, $n = 45$ ( 1 to 45)

Number of numbers to select, $r=5$

Here, order of selection is not important. So, when the order is not important or it doesn't matter how we select objects, then we use combinations formula.

Number of ways of selecting 'r' distinct objects from a collection of 'n' different objects is given as:

$_{r}^{n}\textrm{C}=\frac{n!}{r!(n-r)!}$

Here, 'n' = 45, 'r' = 5. Therefore,

$_{5}^{45}\textrm{C}=\frac{45!}{5!(45-5)!}\\_{5}^{45}\textrm{C}=\frac{45!}{5!(40)!}\\_{5}^{45}\textrm{C}=\frac{45\times 44\times 43\times 42\times 41\times 40!}{5\times 4\times 3\times 2\times 1\times (40)!}\\\textrm{Cancelling out same terms, we get:}\\_{5}^{45}\textrm{C}=\frac{45\times 44\times 43\times 42\times 41}{5\times 4\times 3\times 2\times 1}=12,21,759$

Therefore, the different selections possible are 12,21,759.