# Thread: Permutation and Combinations

1. this is a real simple question if you think about it, but while i was oing the ther permutation questions, i cant seem to pull of an answer to this.. can anyone help me out? thank you five cards bearing the numbers 3,5,6,7, and 8 are given
a) how many numbers greater than 5000 can be formed??
b)how many even numbers greater than 50000 be formed?
c)how many odd numbers less than 400 can be formed?  2.

3. Originally Posted by Heinsbergrelatz
this is a real simple question if you think about it, but while i was oing the ther permutation questions, i cant seem to pull of an answer to this.. can anyone help me out? thank you five cards bearing the numbers 3,5,6,7, and 8 are given
a) how many numbers greater than 5000 can be formed??
1296 Originally Posted by Heinsbergrelatz
b)how many even numbers greater than 50000 be formed?
84 Originally Posted by Heinsbergrelatz
c)how many odd numbers less than 400 can be formed?
6

This sounds rather like a homework problem. I will not give you the rationale.
You should be able to figure out for yourself the reason for the answers given.  4. i knew someone would say it, and indeed is not a homework. im studying for my exams, and just going through all exercises in chapter 7 (which is permutations and comb..)

thank you for your help, but im afraid the answers you have given are wrong.
first of all i have solved the first question and the first question is supposed to be like this; 5!+4*4*3*2 giving 216 not 1296, i have no idea where you got that, but well...yea  5. Originally Posted by Heinsbergrelatz
i knew someone would say it, and indeed is not a homework. im studying for my exams, and just going through all exercises in chapter 7 (which is permutations and comb..)

thank you for your help, but im afraid the answers you have given are wrong.
first of all i have solved the first question and the first question is supposed to be like this; 5!+4*4*3*2 giving 216 not 1296, i have no idea where you got that, but well...yea
You are right. I slipped a decimal and the answer is 120 + 96 = 216

Not 1200 + 96 = 1296  6. also your b) and c) also does not correspond to my working..... im not sure who is right but my my answer also matches with the answer key.  7. So, for those of us for whom it's been more than a decade since a math class, how does one arrive at the above? What are the basic steps involved? I fairly competent with numbers, but the language of math is one I haven't had much opportunity to practice with concepts such as permutations and combinations as of late.  8. this is under Binomial theorem. im sure you have heard of that, its around the level of 15~16 years old. a chapter before you enter calculus at least for my course.  9. The basic steps are about the same. I always forget which is which, but in his example, it's a matter of how many choices you have for each slot. 5 for the first slot, times 4 for the second (you have one less choice since one of the numbers was already chosen) times 3 for the third, 2 for the fourth and 1 for the fifth. 5! different ways to arrange 5 numbers. and then, if you consider combinations that dont include all of the numbers, then you get if I'm not mistaken. most of combinations and permutations, and I vaguely remember the notation but don't know how to reproduce it in a message  10. Originally Posted by inow
So, for those of us for whom it's been more than a decade since a math class, how does one arrive at the above? What are the basic steps involved? I fairly competent with numbers, but the language of math is one I haven't had much opportunity to practice with concepts such as permutations and combinations as of late.
For the first one, you first need to determine how many permutations of the 5 numbers are possible, as any 5 digit number is > 5000. This is done simply with 5!. You also need to determine how many four digit numbers are greater than 5000. For this, you know that you can take any of the cards given except 3, place three more cards after it, and it will be greater than 5000. So, there are 4 possible choices for the first card, then 4 more for the second, 3 for the third, and 2 for the fourth. This gives: The second part requires the number be even, so there are two choices for the final card (6 & 8). Then the numbers for the first four cards (we're doing 5-digit numbers > 5000 first) can be found in 4! ways. For 4-digit numbers > 5000 the final digit must still be a 6 or 8. Additionally, the first digit can not be 3. So there are 3 possible numbers for the first digit (5, 7, and either the 6 or 8, whichever one is not used as the final digit). Then there are 3 possible digits for the second digit (3 is now a valid choice) and 2 choices for the third number. This gives us For the final section, you need to determine how many 1-digit, 2-digit, and 3-digit (less than 400) odd numbers there are. So, one digit is simple: there are 3 odd numbers less than 300. For two-digit, there are 3 choices for the final digit, and 4 for the first digit (digit in the ten's place). For 3-digit odd numbers < 400, there is only 1-choice for the first digit (hundreds place) and that is 3. There are then 2 digits possible for the final digit (since 3 must be in the hundreds place). That leaves 3 possibilities for the middle digit. This gives us the following   Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement