A Group consists of 4 couples in which each of the 4 persons have one wife each. In how many ways could they be arranged in a straight line such that the men and women occupy alternate positions?

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.

He can arrange his schedule in  8 P 6 = 20160 ways

13 P 3= 13!/10! = 1716

This is the number of permutations of five things taken all at a time.

Therefore, answer = 5 P 5 = 120 ways

Since,each member of the league must meet every other member of the league.If they only played each other once,there would be  8 C 2 games.Since,each pairing of teams will occur three times,the answer will be triple.

Therfore, 8 C 2 * 3=84

 Since 8 does not occur in 1000, we have to count the number of times 8 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, where  0 ? x , y , z ? 9 .

Let us first count the numbers in which 8 occurs exactly once.

Since 8 can occur atone place in  3 C 1ways. There are  3 x 9 x 9 such numbers.

Next, 8 can occur in exactly two places in  3 C 2 × 9 such numbers. Lastly, 8 can occur in all three digits in one number only.

Hence, the number of times 8 occur is 

  1 × 3 × 9 2 + 2 × 3 × 9 + 3 × 1 = 300

Given G + B= 41 and B = G-1
Hence, G = 21 and B = 20
Now we have 2 options,
1G and 3M
(or)
3G and 1M
(2G and 2M or 0G and 4M or 4G and oM are not allowed),

Total : C 1 21 × C 3 20 + C 3 21 × C 1 20= 50540 ways.

The given digits are 1, 2, 3, 5, 7, 9  

Number of even numbers = ?P? = 60.

The first, the seventh, the ninth and the tenth letters of the word RECREATIONAL are R, T, O and N respectively. Meaningful word from these letters is only TORN. The third letter of the word is ?R?.

When R and W are the first and last letters of all the words then we can arrange them in 5!ways. Similarly When W and R are the first and last letters of the words then the remaining letters can be arrange in 5! ways.

Thus the total number of permutations = 2 x 5!  = 2 x 120 = 240