ab | ba | ca | da |
ac | bc | cb | db |
ad | bd | cd | dc |
8P3 | = | 8· 7· 6 |
= | 56· 6 | |
= | 50· 6 + 6· 6 | |
= | 336 |
8! 5! | = | 8· 7· 6 |
nPk | = | n! (n − k)! | . . . . . . . . . . . .(1) |
Solution. | 10P4 | = | 10! 6! |
Solution. nPn | = | n! (n − n)! | = | n! 0! | = | n! 1 | = n! |
xyz, xzy, yxz, yzx, zxy, zyx.
4! = 1· 2· 3· 4 = 24
7! = 5,040
Set b as the first letter, and permute the remaining 6. Therefore, there are 6! such arrangements.
The same. 6!.
Begin by permuting the 5 things -- num, b, e, r, s. They will have 5! permutations. But in each one of them, there are 3! rearrangements of num. Consequently, the total number of arrangements in which n, u, and m are together, is 3!· 5! = 6· 120 = 720.
5! = 120
Since 0 cannot be first, remove it. Then there will be 4 ways to choose the first digit. Now replace 0. It will now be one of 4 remaining digits. Therefore, there will be 4 ways to fill the second spot, 3 ways to fill the third, and so on. The total number of 5-digit numbers, then, is 4· 4! = 4· 24 = 96.
Again, 0 cannot be first, so remove it. Since the number must be odd, it must end in either 1 or 3. Place 1, then, in the last position. _ _ _ _ 1. Therefore, for the first position, we may choose either 2, 3, or 4, so that there are 3 ways to choose the first digit. Now replace 0. Hence, there will be 3 ways to choose the second position, 2 ways to choose the third, and 1 way to choose the fourth. Therefore, the total number of odd numbers that end in 1, is 3· 3· 2· 1 = 18. The same analysis holds if we place 3 in the last position, so that the total number of odd numbers is 2· 18 = 36.
The number of permutations of 5 different things taken 3 at a time.
a) nPk | n! (n − k)! | b) 12P7 | 12! 5! |
c) 8P2 | 8! 6! | d) mP0 | m! m! |