How many four digit even numbers can be formed using the digits {2, 7, 5, 3, 9, 1} ?

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.

 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

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

    4 ! × 4 P 4 + 4 ! × 4 P 4 = 2 × 576 = 1152

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

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

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

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.

Required number of ways = 7 C 5 × 3 C 2 = 63