Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

If all the 6 digits are all non-zero 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!
Thanks 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!)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question