The required "option d) 4845" is correct.
Step-by-step explanation:
The number of students in the class = 20
Students to pick up books for the class = 4
Here n = 20 and r = 4
To find, the total number of combinations of 4 students can he choose = ?
∴ [tex]^{20}C_4[/tex]
[tex]=\dfrac{20!}{4!(20-4)!}[/tex]
Using the formula,
[tex]^{n}C_r =\dfrac{n!}{r!(n-r)!}[/tex]
[tex]=\dfrac{20\times 19\times 18 \times 17 \times 16!}{24\times 16!}[/tex]
[tex]=\dfrac{20\times 19\times 3 \times 17}{4}[/tex]
[tex]=5\times 19\times 51=95\times 51[/tex]
= 4845
The total number of combinations of 4 students can he choose = 4845
Hence, the required "option d) 4845" is correct.
