Answer:
[tex]5P3*6C4=60*15=900[/tex]
Step-by-step explanation:
[tex]5P3[/tex] refers to the permutations of 5 items taken 3 at a time. To evaluate this, we use factorials as follows;
[tex]5P3=\frac{5!}{(5-3)!}[/tex]
The factorial of an integer n is evaluated as;
[tex]n!=n*(n-1)*(n-2)*(n-3).......3*2*1[/tex]
Using this concept, the above expression can now be simplified as follows;
[tex]5P3=\frac{5!}{2!}=\frac{5*4*3*2!}{2!}=5*4*3=60[/tex]
Therefore, the permutations of 5 items taken 3 at a time is 60.
The next expression, [tex]6C4[/tex] refers to the combinations of 6 items taken 4 at a time. The simplification utilizes similar concepts of permutations since we shall be involving factorials;
[tex]6C4=\frac{6!}{4!(6-4)!}=\frac{6!}{4!*2!}=\frac{6*5*4!}{4!*2*1}=\frac{6*5}{2*1}=15[/tex]
Therefore, the combinations of 6 items taken 4 at a time is 15.
The final step is to evaluate the product;
[tex]5P3*6C4=60*15=900[/tex]