Is it "A"
360 ways

Na....I got the same...
But the OA is B) 4096

The given solution is like this:
"r things can be given to n persons in n^r ways"
Hence 6 rings can be worn on 4 fingers in 4^6 ways = 4096 ways

OR

(4C1)^6...

Any explanations?

rishi235 wrote:Na....I got the same...
But the OA is B) 4096

The given solution is like this:
"r things can be given to n persons in n^r ways"
Hence 6 rings can be worn on 4 fingers in 4^6 ways = 4096 ways

OR

(4C1)^6...

Any explanations?

for me its B 4^6

its a common formula inn P&C

The number of ways in which u can arrange r things to n persons/things is n^r

heres how it works'

consider there are r boxes and n items. the first box can be filled in n ways and the second can be filled in n ways as well ( even thought the first is filled with n ways ,, we can use the same item for filling the second box).

thus the rth box can be filled in n ways as well
Now all the r boxes together can be filled in
n*n*n*n................r times which is n^r ways

use the same logic here

u will get

4*4*4............ 6 times which is 4^6

hope i made is clear...do let me know if u have any doubts...

Got it Sudhir...Thanks

Just cam across another similar problem

Q: In how many ways can u put 9 coins of DIFFERENT values into 2 packs

Ans will straight be 2^9...

But when we modify this questions as:
In how many ways can u put 9 coins of 6 DIFFERENT values into 2 packs..

Ans: 2^9 / 3!....as 3 coins are identical...
Is this the right answer for the modified problem?
Or will d answer b 2^9/2^3?

Kindly advice.... Thanks

rishi235 wrote:Got it Sudhir...Thanks

Just cam across another similar problem

Q: In how many ways can u put 9 coins of DIFFERENT values into 2 packs

Ans will straight be 2^9...

But when we modify this questions as:
In how many ways can u put 9 coins of 6 DIFFERENT values into 2 packs..

Ans: 2^9 / 3!....as 3 coins are identical...
Is this the right answer for the modified problem?
Or will d answer b 2^9/2^3?

Kindly advice.... Thanks

You can't answer unless you know the specific duplications.

For example, let's say you have 1c, 5c, 10c, 25c, 50c and 100c coins - that's 6 different values.

However, we could have 9 coins that are:

1c 1c 1c 1c 5c 10c 25c 50c 100c

or

1c 1c 5c 5c 10c 10c 25c 50c 100c

Each of those sets would give a different answer to the question (in the first case, we'd factor out 4!; in the second, we'd factor out 2!2!2!).

Thanks Stuart & Sudhir...but i've got another doubt now

Q 1 - If v want to arrange 1,2,3,4 in 3 places
1) Without repetition - Ans = 4P3 &
2) With repetition - Ans = 4^3 ??? Is this right?

I'm posting another similar problem from the same P&C book. I'm posting it in this thread as it is similar to the 1st problem and it will b easy for all to relate the difference...

Q 2 - How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels? (REPETITION IS ALLOWED)

No options available....
I didnt follow the explanation given in the book...
It mentions r^n instead of n^r, which sudhir mentioned in his previous post..
Kindly explain this discrepancy ....

Thanks...

rishi235 wrote:Thanks Stuart & Sudhir...but i've got another doubt now

Q 1 - If v want to arrange 1,2,3,4 in 3 places
1) Without repetition - Ans = 4P3 &
2) With repetition - Ans = 4^3 ??? Is this right?


Thanks...

rishi ur right on this one..but dont confuse urself too much with P's and z^3....heres how its works...

u have 3 places and 4 numbers..

without repetition
first place can be filled from 4 numbers in 4 ways
second place in 3 ways ( one number is already used in the first place)
3rd place in 2 ways ( 2 numbers already used)

thus its 4*3*2 = 24 or 4P3...

with repetition

first place in 4 ways
2nd place also in 4 ways ( u can take the same number again ..so u again have 4 numbers to choose from )
3rd also in 4 ways

thus its 4*4*4 = 64 or 4^3 ...

Hope it helps u..do let us know if have any doubts

Thanks a lot ....U have really made it look simple...
I always got confused with such problems...

Now with this understanding...i cud understand & solve the 2nd problem too

Thanks once again

Sorry, could we take another look at the original problem?

I think the answer is A - my own math aside, heres an example from MGMAT cominatorics strategy:

If there are 7 people in a room, but only 4 chairs, how many different seating arrangements are possibe;

Answer; 7!/3! = 840. So, we have more 'rings', the same amount of 'fingers', but far fewer combos. Plus, 4,000+ just seems counterintuiitive to me.

Not sure how the answer isn't 6!/2~ = 360 unless I'm misreading the original problem...thoughts?

BILL wrote:Sorry, could we take another look at the original problem?

I think the answer is A - my own math aside, heres an example from MGMAT cominatorics strategy:

If there are 7 people in a room, but only 4 chairs, how many different seating arrangements are possibe;

Answer; 7!/3! = 840. So, we have more 'rings', the same amount of 'fingers', but far fewer combos. Plus, 4,000+ just seems counterintuiitive to me.

Not sure how the answer isn't 6!/2~ = 360 unless I'm misreading the original problem...thoughts?

Because nothing in the question says there's a maximum number of rings per finger, where in the chair/people question it's 1 person per chair.