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

 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

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

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