Hin Dus Tan

Probability Level 4

The number of permutations of the letters of the word HINDUSTAN such that neither the pattern HIN nor DUS nor TAN appears is $x.$ Find the digit sum of $x.$

Details and assumptions

The digit sum of 34566 is $3 + 4 + 5 + 6 + 6=24.$

The answer is 30.

1 solution

Rajen Kapur
Aug 21, 2014

Anybody getting the answer as 168942 will also report 30 as digit sum. There are 2 N's hence permutations of HINDUSTAN are 9!/2, and not 9!, as claimed by Sandeep.

If somebody will go through this way then also he will report 30 as answer but this solution is absolutely wrong

Total no of permutations $9!$ Number of permutations having all three patterns $3!$ Number of permutations having any 2 patterns $³c_{2}\times{5!}$ Number of permutations having any one pattern $³c_{1}\times{7!}$

Therefore number of required permutations :

$9!-{(³c_{2}5! + ³c_{1}7! + 3!)} = 347394$

3+4+7+3+9+4=30

BUT

Correct answer should be $\boxed{169194}$

Kïñshük Sïñgh - 6 years, 9 months ago

I agree correct answer should be 169194.

Ronak Agarwal - 6 years, 9 months ago

Yes, the correct answer is 169194

Karan Siwach - 6 years, 9 months ago

Why is the answer 169194? Can u provide a solution pls?

Jackal Jim - 6 years, 9 months ago

$\frac{9!}{2}-(7!+7!+\frac{7!}{2})+(3\times{5!}-3!)$

$181440-12600+354=169194$

I think this will help you

Kïñshük Sïñgh - 6 years, 9 months ago

@Kïñshük Sïñgh – I too have done the same way.

Ronak Agarwal - 6 years, 9 months ago

@Kïñshük Sïñgh – I did the 9!/2 - [7! + 7! +7!/2] part.. how do u explain the remaining part?

Chinmay Raut - 6 years, 9 months ago

@Chinmay Raut – $Lets\quad understand\quad this\quad with\quad an\quad example,\\ first\quad club\quad HIN\quad \\ 1.\quad HIN\_ \quad \_ \quad \_ \quad \_ \quad \_ \quad \_ \quad \\ \quad \quad DUS\_ \quad \_ HIN\_ \quad (one\quad of\quad the\quad arrangement)\\ ways\quad of\quad arranging\quad =\quad 7!\quad \\ \\ 2.\quad now,\quad club\quad DUS\quad \\ \quad \quad DUS\_ \quad \_ \quad \_ \quad \_ \quad \_ \quad \_ \quad \\ \quad \quad DUS\_ \quad \_ HIN\_ \quad (one\quad of\quad the\quad arrangement)\\ \quad \quad ways\quad =\quad \frac { 7! }{ 2 } \quad \\ therefore\quad there\quad will\quad be\quad repetation\quad of\quad some\quad cases\quad like\quad these\quad which\quad we\quad have\quad counted\quad in\quad 1.\quad \\ thats\quad why\quad we\quad have\quad to\quad add\quad 3\times 5!\quad and\quad one\quad case\quad which\quad will\quad be\quad counted\quad in\quad all\quad the\quad arrangements\quad (HINDUSTAN),\quad therefore\quad we\quad have\quad to\quad substract\quad it\quad \\ \\ therefore\quad our\quad required\quad answer\quad will\quad be\quad \\ \frac { 9! }{ 2 } -(7!+7!+\frac { 7! }{ 2 } )+(3\times 5!-3!)=169194$

Kïñshük Sïñgh - 6 years, 9 months ago

Hin Dus Tan

The answer is 30.

1 solution

3+4+7+3+9+4=30

BUT

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