= 15504
6P4 = 6! / (6-4)! = 360
P(4,3)= = 24
Three digit number will have unit?s, ten?s and hundred?s place.
Out of 5 given digits any one can take the unit?s place.
This can be done in 5 ways. ... (i)
After filling the unit?s place, any of the four remaining digits can take the ten?s place.
This can be done in 4 ways. ... (ii)
After filling in ten?s place, hundred?s place can be filled from any of the three remaining digits.
This can be done in 3 ways. ... (iii)
So,by counting principle, the number of 3 digit numbers = 5x 4 x 3 = 60
Required number of ways =
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =7!/2!= 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 3!/5!= 20 ways.
Required number of ways = (2520 x 20) = 50400.
Since it does not matter what order the committee members are chosen in, the combination formula is used.
Committees are always a combination unless the problem states that someone like a president has higher hierarchy over another person. If the committee is ordered, then it is a permutation.
C(17,7)= 19,448
When a hand of cards is dealt, the order of the cards does not matter. If you are dealt two kings, it does not matter if the two kings came with the first two cards or the last two cards. Thus cards are combinations. There are 52 cards in a deck and we want to know how many different ways we can put them in groups of five at a time when order does not matter. The combination formula is used.
C(52,5) = 2,598,960
Since order does not matter it is a combination.
The word AND means multiply.
Given 4 basketball, 3 volleyball, 2 soccer.
We want 2 basketball games and 1 other event. There are 5 choices left.
C(n,r)
C(How many do you have, How many do you want)
C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1)
C(4,2) x C(5,1) = (6)(5) = 30
Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.
There are 5 slots.
__ __ __ __ __
The first slot must be a four. There are 4 ways to put a four in the first slot.
There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.
(4)(4)(4)(4)(4) = 1024
Therefore there are 1024 different ways to produce the desired hand of cards.
There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.
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