COMBINATORICS!!!

For students of class 6-8 (age 12 to 14)
aritra barua
Posts: 48
Joined: Sun Dec 11, 2016 2:01 pm

COMBINATORICS!!!

Unread post by aritra barua » Mon Jan 23, 2017 7:45 am

Observe the following cases:1222111,3334447,6669222;in each of the cases, the used digits appear at least thrice.Find the number of such 7 digit numbers with this aforementioned property.... :(

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Tasnood
Posts: 11
Joined: Tue Jan 06, 2015 1:46 pm

Re: COMBINATORICS!!!

Unread post by Tasnood » Tue Jan 24, 2017 10:47 pm

Not understand. In the second case, I see 7 has appeared once. So, where each digiit is appearing thrice (at least)?

aritra barua
Posts: 48
Joined: Sun Dec 11, 2016 2:01 pm

Re: COMBINATORICS!!!

Unread post by aritra barua » Wed Jan 25, 2017 7:19 pm

Well,those were typing mistakes.Avoid those :)

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Tasnood
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Re: COMBINATORICS!!!

Unread post by Tasnood » Thu Jan 26, 2017 8:47 pm

We assume that a number of 7 digits where at least 3 digits are same. As for example, if we take such a number, n = 111222 and the last digit will be 1 or 2. So, total combination: $\frac{7!}{3!4!}$ for the last digit 1 and same for 2. So, like (1,2), there are {(1,3),(1,4),(1,5),...,(1,9)} and 70 combination for each. If we take {(2,3),...,(2,9)} 70 will be each. And, for (8,9) the answer is same. so, total = $70\cdot{(8+7+...+1)}$ = $70\cdot36$ = 2520.

Now, we think about 0. For n = 1111000, number of combination = $\frac{6!}{3!3!}$ = 20 and for n = 1110000, combination = $\frac{6!}{2!4!}$ = 15, because a 1 must be at first place.
Same for {2,3,...,9} and result in this case = $9\cdot{(20+15)}$ = $9\cdot35$ = 315.

The third case, if all digits are same. Then, there will be 9 digits.
In total = 2520 + 315 + 9 = 2844; the answer we desired.

aritra barua
Posts: 48
Joined: Sun Dec 11, 2016 2:01 pm

Re: COMBINATORICS!!!

Unread post by aritra barua » Sat Jan 28, 2017 1:49 pm

Correct!My solution is more or less as yours too..... :D

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