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Permutation and Combination Help

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Intern
Joined: 28 Feb 2011
Posts: 32

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22 Apr 2011, 09:14
Hi Guys,

I came across this problem in tutorvista site.

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Find the number of words which have atleast one letter repeated.

I followed a different approach for solving and got a diiferent /wrong answer. Here's the appraoch that i followed.

Since 5 letters need to formed with atleast 1 word repeated, we have the following 3 cases:

CASE 1) xx123, where 1 word (XX) is repeated and 123 r the remaining 3 words

case 2) xxyy4, where 2 words r repeated (XX and YY) and 4 is the last word

CASE 3) xxxxx where all the word r repeated

CASE 1:

xx123

1st letter can be selected from 10 leeters in 10c1ways, since the second letter is repeated, there is only 1 way to do this. similarly , 3rd, 4th and 5th posotion can be filled in 9c1, 8c1 and 7c1 ways respectively.

so the total is : 10c1* 1* 9c1*8c1*7c1 ways ..further each of these words can be arranged in 5!/2! ways

So toatal is : 10c1* 1* 9c1*8c1*7c1 *5!/2! = 302400

CASE 2:

xxyy4

No of five letter words that can be formed = 10c1 *1 * 9c1 *1 *8c1 ways..furhter teses can be arranged in 5!/2!2! ways

So total number of 5 letters that can be formed, where two letters r repeated r = 10c1 *1 * 9c1 *1 *8c1 *5!/2!2! = 21600 ways

CASE 3)

xxxxx

10c1*1*1*1*1 ways . further this can be arraged in 1 ways only
So total number of 5 letter word that can be formed where alll leters r repeated is 10

TOTAL = CASE 1+CASE2+CASE3

302400 +21600+10

Can somebody please tell me where have i gone wrong?

Thanks,
Anu
Retired Moderator
Joined: 20 Dec 2010
Posts: 1462
Re: Permutation and Combination Help  [#permalink]

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22 Apr 2011, 09:25
1
anuu wrote:
Hi Guys,

I came across this problem in tutorvista site.

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Find the number of words which have atleast one letter repeated.

I followed a different approach for solving and got a diiferent /wrong answer. Here's the appraoch that i followed.

Since 5 letters need to formed with atleast 1 word repeated, we have the following 3 cases:

CASE 1) xx123, where 1 word (XX) is repeated and 123 r the remaining 3 words

case 2) xxyy4, where 2 words r repeated (XX and YY) and 4 is the last word

CASE 3) xxxxx where all the word r repeated

CASE 1:

xx123

1st letter can be selected from 10 leeters in 10c1ways, since the second letter is repeated, there is only 1 way to do this. similarly , 3rd, 4th and 5th posotion can be filled in 9c1, 8c1 and 7c1 ways respectively.

so the total is : 10c1* 1* 9c1*8c1*7c1 ways ..further each of these words can be arranged in 5!/2! ways

So toatal is : 10c1* 1* 9c1*8c1*7c1 *5!/2! = 302400

CASE 2:

xxyy4

No of five letter words that can be formed = 10c1 *1 * 9c1 *1 *8c1 ways..furhter teses can be arranged in 5!/2!2! ways

So total number of 5 letters that can be formed, where two letters r repeated r = 10c1 *1 * 9c1 *1 *8c1 *5!/2!2! = 21600 ways

CASE 3)

xxxxx

10c1*1*1*1*1 ways . further this can be arraged in 1 ways only
So total number of 5 letter word that can be formed where alll leters r repeated is 10

TOTAL = CASE 1+CASE2+CASE3

302400 +21600+10

Can somebody please tell me where have i gone wrong?

Thanks,
Anu

It's sometimes easy to solve it using another approach;

Total words that can be formed with repetition: $$(10)^5$$

Total words with no repetition= $$P^{10}_{5}=\frac{10!}{5!}$$

Total words with at least one repetition $$(10)^5 - \frac{10!}{5!}$$
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10793
Location: Pune, India
Re: Permutation and Combination Help  [#permalink]

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22 Apr 2011, 14:13
anuu wrote:

Can somebody please tell me where have i gone wrong?

Thanks,
Anu

What about 'xxx12' or 'xxxyy' or 'xxxx1'? (I am assuming x and y are different)
The 'atleast 1' questions are easier if done using the method shown by fluke.
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Karishma
Veritas Prep GMAT Instructor

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Joined: 09 Sep 2013
Posts: 15625
Re: Permutation and Combination Help  [#permalink]

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09 May 2017, 18:12
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Re: Permutation and Combination Help   [#permalink] 09 May 2017, 18:12