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Source: — Problem Solving |
Newbie | Next Rank: 10 Posts
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hi

by sudhir_83k » Tue Jul 29, 2008 12:33 am
is the answer to this is 15C5.
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by parallel_chase » Tue Jul 29, 2008 12:53 am
My answer is 13, whats the OA?

Here's the explanation.

We have 5 distinct letters to choose from. A,B,C,D,E

All of these letters can be chosen= 1 ways.

All A's can be chosen = 1ways

All B's can be chosen = 1*4C1=4ways

All C's can be chosen = 1*4C2=3ways

All D's can be chosen = 1*4C3=4ways



Therefore total number of ways =1+1+4+3+4 = 13 ways
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by sudhir3127 » Tue Jul 29, 2008 6:29 am
Hi All... thanks for the discussion,,

the answer is 71.

here it goes...

Number of ways of selecting 5 different letters = 5C5 = 1 way

Number of ways to select 2 similar and 3 different letter = 4C1* 4C3=16

Number of ways of selecting 2 similar + 2 more similar letter and 1 different letter = 4C2* 3C1= 18

Number of ways to select 3 similar and 2 different letter = 3C1* 4C2= 18

Number of ways to select 3 similar and another 2 other similar = 3C1*3C1=9

Number of ways to select 4 similar and 1 different letter = 2C1*4C1=8

ways of selecting 5 similar letters = 1

total ways = 1+16+18+18+9+8+1= 71

Hope its clear.. please let me know if u have any doubts...

P.S How do u guys rate such questions,,. just wanted to understand the level of toughness of such questions..
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by preetha_85 » Tue Jul 29, 2008 10:04 am
Hi

If we have 2 select 2 similar nd 3 diff letters will it not be :
4C3*(5C2+4C2+3C2+2C2)

Nd the same way for all the other case also....
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by pepeprepa » Wed Jul 30, 2008 12:15 am
Sudhir, I find this way of solving pretty long.
I did the same but it is
1) long
2) you can make many errors if there is a lack of concentration
Is the correction of the exercice faster?
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