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22 Apr 2011, 10:14
Kudos
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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
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
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22 Apr 2011, 10:25
Kudos
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anuu
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 \((10)^5 - \frac{10!}{5!}\)
22 Apr 2011, 15:13
Kudos
Bookmarks
anuu
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.
