There are 12 people, so this is our n value.

So, 12C21= 66

Two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4. 

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

Entity is a real world object

LDI is the capacity to change the conceptual schema without having to change external schemas or application programs.

There are 5 slots.

                   __ __ __ __ __

The first slot must be a four. There are 4 ways to put a four in the first slot.

There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.

(4)(4)(4)(4)(4) = 1024

Therefore there are 1024 different ways to produce the desired hand of cards.

i=j/m

Maturity value
= PV(1 + i)^n

Term
= 5 years - 21 months=
3.25 years

Price paid
= FV(1+ i )^-n

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