Please solve

1) Combination of 2 men and 4 women in 4C2 x  6C4 ways = 90.

Combination of 3 men and 6 women in 4C3 x 6C6 ways = 4. 

Total 94.

<br>

..

2) Sum of n terms = n/2(2a +nd -d) which is the standard form.

Given expression of sum of n terms is 2n^2+n. Expressing it in standard form we get, n(2n+1) i.e. n/2(4n+2).

Comparing nth term in both the standard form and modified expression, we find that for the only n term within brackets is 4n, equating it to nd, we get d=4.

Now, difference of tenth and first term is a+9d-a i.e. 9d =36.

3) 8 people can be selected out of 15 to sit in one round table in 15C8 ways. Remaining 7 people will go to the other table automatically.

Now, 8 persons can arrange amongst themselves in (8-1)! ways on one table and 7 persons can arrange amongst themselves on the other table in (7-1)! ways.

So total number of ways are 15C8*7!*6!.

4) Units digit can be filled up only by 0, 2 or 4. 

Now, fixing zero in unit's place, remaining 2 digits can be selected in 5P2 ways i.e. 20 ways.

Now, fixing 2 in unit's place, remaining 2 digits can be selected in 5P2 ways subtracted by cases in which zero are in the hundreth's place which can be selected in 4P1 ways giving (5P2-4P1) i.e. 16 ways.

Now, fixing 2 in unit's place, remaining 2 digits can be selected in 5P2 ways subtracted by cases in which zero are in the hundreth's place which can be selected in 4P1 ways giving (5P2-4P1) i.e. 16 ways.

So total = 52.

5) interest given by PRT/100. Here, P is x and y for the two sums.

So, equating both interests, x*6*7/100=y*9+10/100.

Solving, x/y=90/42 i.e.15:7

many many thanks to u.........