permutation nd combination

1. the number to be divisible by 4 must end with 24

the rest 2 digits have to be selected from 1,3,5.

that can be done by ³C₂ i.e 3 ways.

hence it is the answer.

2. the number must be less than 10000.

thus it can be one digit, two digit, three digit or four digit number.

all 10 one digit have no repitition.

in 2 dig numbers, the tens place can be filled in by 9 ways(except 0) and the ones place can be filled by 9 digits again(except the digit already used). so in 9X9=81 ways.

in 3 dig number, the hundreds tens and ones place can be respectively filled in 9,9,8 ways.
thus no of ways=9X9X8=648

in 4 digits similarly the filling up can take place in 9X9X8X7 ways i.e. 4536 ways.

the sum 10+81+648+4536 gives the answer!(too lazy to add up :()

3. well the setup of this question is tricky.

there can be any number or letters in the word.

and as nothing is mentioned, we take that repitition is allowed.

that gives an answer infinite

[3]

this time i want to say CHECK QUESTION!

and if the question was. no of words with 8 letters, no repitition,begins and ends with consonant then ways=3X6X5X4X3X2X1X4= 8640!!

if repititin was allowed then answer=4X8X8X8X8X8X8X4=2X8⁷

@subhomoy.. the queston 1 is solved incorrectly.. you have missed 32 ...

So the answer will be double of what you have got..

oops!! yes that i missed out!! :P

so corrected Q1)

the last 2 digits can be either 12, 24, 32 or 52 (why?)

considering the case of 24 only we get ³C₂.

Similarly, we will get the same result for 12, 32 and 52...

thus, No of such words=4X³C₂=12!!

Nishant bhaiya u also missed 12, 52 as i did! hehehe! ;)

lol :P