#1 There are two 8's, so while there are 6! permutations of the six digits, in any one of them, the two 8's can be swapped, and it's still the same arrangements. Same for the 5's. So, there are only 6!/(2!2!) = 180 distinguishable arrangements.
The same caveat applies to all the other problems. You need review your text section and rethink your answers.
Am I correct? Please show me how if I am wrong
1.How many 6 digit numbers can be made using each of the digits {8,8,9,5,5,6} exactly once? 6!=702
2.How many more arrangements are there from all of the letters of the word MARMALADE than there are from all of the letters of the word MARINADE? 9!-8!=322560
3.Using all of the digits {1,1,1,2,2,3,3}, how many positive seven-digit integers can be written that begin and end in 3? 3x5x5x5x5x5x3=28125
4.In how many ways can 3 yellow beads, 4 blue beads and 2 black beads be strung along a straight wire if all of the beads are identical except for color? (Don't know)
5.In how many ways can all of the letters of the word CANADA be arranged? Since there is 3 a's it is divied by 3! 6!/3!=120
6.In how many ways can all of the letters of the word MOOSOMIN be arranged? 2 m's and 3 o's 8!/2!x3!
7.Using all of the digits {1,1,1,2,2,3,3}, how many positive seven-digit integers can be written that begin and end in 1? 1x5x5x5x5x5x1=3125
8.In how many different orders can a soccer team win 4 games and lose 3 games in a best-of-seven playoff series? 7!/4!x3!=20160
3 answers
1. 6!/(2!2!)
2.(9!/(2!3!)-8!/2!
3.3x(5!/3!2!)x3
4.9!/3!4!2!
5. 3! 6!/3!
6.8!/2!x3!
7.1x(5!/3!2!)X1
8.7!/4!3!
2.(9!/(2!3!)-8!/2!
3.3x(5!/3!2!)x3
4.9!/3!4!2!
5. 3! 6!/3!
6.8!/2!x3!
7.1x(5!/3!2!)X1
8.7!/4!3!
1. 180
2. 10080
3. 94. 0
4. 2,903,040
5. 120
6. 3360
7. 10
8. 35
2. 10080
3. 94. 0
4. 2,903,040
5. 120
6. 3360
7. 10
8. 35
0 answers