Since,each member of the league must meet every other member of the league.If they only played each other once,there would be games.Since,each pairing of teams will occur three times,the answer will be triple.
Therfore, =84
There are ways of selecting two men, and ways of selecting a woman.Since each position in the committee is different,arrange the three people in 3! ways.
So,
First we need to choose two vowels and then three consonants . Now that we have 5 letters required to make the word,arrange them in 5! ways.
So,
If Jason is on th ecouncil,this reduces the selction pool to only 13 people,out of which we still need to select 4.
So, = 715
The total possible cases would be a 5 card hand with no restrictions : 5
The unwanted cases are:
no queens(out of 48 non-queens cards we get 5)
only 1 queen(out of 4 queens we get 1,and out of 48 non-queens we get 4)
Therefore,
This is the number of permutations of five things taken all at a time.
Therefore, answer = = 120 ways
= 13!/10! = 1716
He can arrange his schedule in = 20160 ways
The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are:
One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.
We can calculate each of these cases and then add them together, or approach this problem from a different angle.
The only case which is left out of the condition is the case where no wrong packages are delivered.
If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.
There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package.
The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
1×1×1=1
Determine the total number of ways the three packages can be delivered.
3×2×1=6
The number of ways at least one house gets the wrong package is:
6?1=5
Therefore there are 5 ways for at least one house to get the wrong package.
Case I : MW MW MW MW
Case II: WM WM WM WM
Let us arrange 4 men in 4! ways, then we arrange 4 women in 4P4 ways at 4 places either left of the men or right of the men. Hence required number of arrangements
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