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29 Mar 2008 16:43:11 IST

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Permutations of letters of word HINDUSTAN such that HIN, DUS nor TAN appears

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The number of permutations of the word HINDUSTAN such that neither the pattern HIN nor DUS nor TAN appears is ?

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Cursed Mask

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Joined: 8 Oct 2007

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29 Mar 2008 16:52:57 IST

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A = total no of permutations = 9!

B = no of permutations in which only one string appears = 3 x {7! - 6}

C = no of permutations in which two strings appear = 3 x {5! - 6}

D = no of permutations in which all three strings appear = 6

Ur answer = A - B - C - D

Cursed Mask

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29 Mar 2008 16:53:53 IST

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I'm sorry...B has been calculated wrongly....but u get the general idea

Cursed Mask

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Posts: 98

29 Mar 2008 16:56:30 IST

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correct answer seems to be

9! - 7! + 5! - 3!

is this correct?

Arnav Guha

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Joined: 22 Feb 2007

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29 Mar 2008 17:49:42 IST

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Total =9!

those in which

HIn Dus TAn not to be rearranged =9!/(3!)³

hence required= 9! - ( 9!/(3!)³)

am i rit ?

Akhil

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Joined: 17 Jan 2008

Posts: 1451

29 Mar 2008 18:20:09 IST

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(S1=one of HIN , DUS TAN appear

HIN) appears:
(HIN)DUSTAN
7 entities arranged in 7! ways
(DUS)HINNTA
7 items, 2 Ns
so 7!/2 ways
(TAN)HINDUS
7! ways

S2: any two pairs together:
now any two of these together :(HIN)(TAN) DUS
5! ways for each combo of 2
so 3*5!

S3: all 3 together
(HIN)(DUS)(TAN) arranged in 3! ways

now we see that required
S1-S2 all individual cases are added
S3 is added thrice and subtracted 3ice
so it has to be added again

So
S=S1-S2+S3
=(7!+7!+7!/2)-(3*(5!))+3!

but required cases= total - S

total=9!/2

so
answer=9!/2-[(7!+7!+7!/2)-(3*(5!))+3!]

edison

Forum Expert

Joined: 19 Oct 2006

Posts: 7537

3 Apr 2008 21:26:54 IST

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very well done akhil

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