Problem Solving

Came across this question in a P & C book (non-GMAT)

In how many ways can u wear 6 rings in 4 fingers of a hand?
A) 360
B) 4096
C) 1296

Pls show your working....

Thanksss

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.