At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Write your question here (Keep it simple and clear to get the best answer) If jun has 12 T-shirt 6 pairs of parts, and 3 pairs of share, how many possibilities can be'drace himself up for the day?

Sagot :

Answer:

the number of possibilities that can be drace himself up for the day is 216

Step-by-step explanation:

Given that

jun has 12 t-shirts,6 pairs of pants and 3 pairs of shoes

We need to find out the possibilities that can be drace himself up for the day

So here

[tex]= 12C_1 \times 6C_1 \times 3C_1\\\\= \frac{12!}{1!(12 - 1)!} \times \frac{6!}{1!(6 - 1)!} \times \frac{3!}{1!(3 - 1)!}\\\\= \frac{12!}{1!(11)!} \times \frac{6!}{1!(5)!} \times \frac{3!}{1!(2)!}\\\\= 12 \times 6 \times 3\\\\= 216[/tex]

Hence, the number of possibilities that can be drace himself up for the day is 216

The same would be considered and relevant too