A community for students.
Here's the question you clicked on:
 0 viewing
hqi1321
 3 years ago
How many 6 digit positive integers are there, consisting of exactly one 3, one 5, two 4s, and the remaining two digits either 7 or 8?
hqi1321
 3 years ago
How many 6 digit positive integers are there, consisting of exactly one 3, one 5, two 4s, and the remaining two digits either 7 or 8?

This Question is Open

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.1If all the 6 digits are all nonzero and distinct, there are 6! possible integers. If there are two identical digits, we need to divide by 2! Don't know if the 7 and 8's mean A. EITHER two 7's OR two 8's,, or B. EACH digit is 7 or 8. Case A: 1. if both are 7's => 6!/(2!2!) 2. if both are 8's => 6!/(2!2!) Total 2*6!/(2!2!) Case B: 1. same digits (7/7 or 8/8) => 2*6!/(2!2!) 2. different digits (7/8 or 8/7) => 6!/2! Total = 2*6!/(2!2!) + 6!/2! =6!

hqi1321
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for your answer! Sorry, it meant each. Paraphrased a little. I'm confused, why do we divide by 2! if there are 2 identical digits? Why do we divide (why isn't it 6!/(2!2!)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.