This question is a combination since order is not important.
Answer = = 126
Two horses A and B, in a race of 6 horses... A has to finish before B
if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!
if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!
if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!
if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4!
if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..
A cannot finish 6th, since he has to be ahead of B
Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360
Selecting zero'A's= 1
Selecting one 'A's = 1
Selecting two 'A's = 1
Selecting three 'A's = 1
Selecting four 'A's = 1
Selecting five 'A's = 1
=> Required number ofways =6
Total number of alphabets = 10
so ways to arrange them = 10!
Then there will be duplicates because 1st S is no different than 2nd S.
we have 4 Is 3 S and 2 Ps
Hence number of arrangements = 10!/4! x 3! x 2! = 12600
fix one person and the brothers B1 P B2 = 2 ways to do so.
other 17 people= 17!
Each person out of 18 can be fixed between the two=18, thus, 2 x 17! x 18=2 x 18!
Since two of the eight first class petty officers are to fill two different offices, we write 56
Then, two of the six second class petty officers are to fill two different offices; thus, we write =30
The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders
There are 7 toppings in total,and by selecting 3,we will make different types of pizza.
This question is a combination since having a different order of toppings will not make a different pizza.
So, =35
since it is a combination = = 2878 x 10^10
If John must be on the committee,we have 11 students remaining,out of which we choose 4. So,
There are 13 spades,we must include 2: 13
There are 13 diamonds,we must include 3:
Since we can't have more than 2 spades and 3 diamonds,the remaining 2 cards must be pulled out from the 26 remaining clubs and hearts :
Therefore, = 7250100
If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.
So, 50
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