A quick refresher for those less familiar with harder permutations.
How many ways can the letters ABC be arranged if no letters are repeated?
We have 3 choices for the 1st position.
We have 2 choices left for the 2nd position. (Because we used 1 of our 3 choices for the first position, leaving us only 2 choices for the 2nd position.)
We have 1 choice left for the last position. (Because we used 2 of our 3 choices for the first two positions, leaving us only 1 choice for the last position.)
Now we multiply the number of choices for each position:
3 * 2 * 1 = 3! = 6 possible arrangements.
So the total number of ways n elements can be arranged is n!
Now a harder question:
How many ways can the letters AAB be arranged if the letter A must appear exactly twice in the arrangement?
Normally, the number of arrangements would be 3!. But we have to account for the repeated A, which will reduce the number of unique arrangements. (If we wrote out all the possible unique arrangements, we'd see that there are only 3: AAB, ABA, and BAA.)
When an element is repeated, use the following formula:
(number of elements!)/(number of repetitions!)
In AAB we have 3 elements and 2 repetitions of the letter A, so we have to divide by 2!:
3!/2!= 3.
Another example:
How many ways can the letters ABBA be arranged?
We have 4 elements to arrange, but because we have 2 A's, we have divide by 2!, and because we have 2 B's, we have to divide again by 2!:
4!/(2! * 2!) = 6.
So moving onto the problem above:
A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?
If all 3 letters have to be included in the code, one letter will need to be used twice. Our options are AABC, BBAC, and CCAB.
Number of ways to arrange AABC: 4!/2! = 12. (We divide by 2! because the A is repeated 2 times.)
Number of ways to arrange BBAC: 4!/2! = 12. (We divide by 2! because the B is repeated 2 times.)
Number of ways to arrange CCAB: 4!/2! = 12. (We divide by 2! because the C is repeated 2 times.)
Adding, we see that there are 12 + 12 + 12 = 36 possible arrangements.
The correct answer is C.
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Are you sure all your actual options are unique, chintudave?
