0 votes
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The number of ways 5 letter be put in 3 letter boxes A,B,C. If letter box A must contain at least 2 letters.

edited | 237 views
0
is answer 10?
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no, it is 131
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madeeasy's answer might be wrong
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x1+x2+x3=5

given condition is that one box contain at least 2 letters which is x1>=2

x1-2>=2-2

x1-2>=0

we can write

x1-2+x2+x3=5

x1+x2+x3=7

using the concept of stars and bars we can determine

||+||+|||=7

9c2=36

what is wrong in this ?

+4

@vijju532

It specifically mentioned box A has atleast 2 letters. Its not right to consider "one box".

Let the 5 letters be named as L1,L2,L3,L4,L5.

Case 1) Put 2 letters in A. Ways of choosing 2 letters from 5 is C(5,2).

No. of ways of putting remaining 3 letters into B and C is $2^3$.

Let A has L1 and L2. So remaining is L3 to L5.

L3 can go to B or C. 2 ways

L4 can go to B or C. 2 ways.

L5 can go to B or C. 2 ways.

So total cases here is $C(5,2)x2^3 =10*8=80$

Case 2)

Put 3 letters in A. Ways of choosing 3 letters from 5 is C(5,3).

No. of ways of putting remaining 2 letters into B and C is $2^2$.

So total cases here is $C(5,3)x2^2 =10*4=40$

Case 3)

Put 4 letters in A. Ways of choosing 4 letters from 5 is C(5,4).

No. of ways of putting remaining 1 letters into B and C is $2^1$.

So total cases here is $C(5,4)x2^1 =5*2=10$

Case 4)

Put 5 letters in A. Ways of choosing 5 letters from 5 is C(5,5).

No. of ways of putting remaining 5 letters into B and C is $2^0$.

So total cases here is $C(5,5)x2^0 =1*1=1$

Sum them up: 80+40+10+1=131

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can we use the concept of onto funtion ?
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@vijju532 I edited the comment. You can check it. Yes concept of on-to can be used.

+2
Another method

$3^{5}$ - ( {when box 1 contain only 1 letter } - {when box 1 contains 0 letter} )

## 2 Answers

+1 vote
The answer depends on weather the letters are same or different. If the letters are same answer is 10, if different answer is 131.
by (241 points)
0
According to que 131 is the right answer.
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You mean $10$ is possible as well?
+1
Yes it is 10 if the letters are indistinguishable.
0 votes

correct answer is 10.

This problem belongs to combinations with repetitions. Try the examples from rosen's book

by (437 points)

+1 vote
3 answers
1
+1 vote
1 answer
2
0 votes
1 answer
3