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From: David C. Ullrich on 24 May 2010 09:02
On Sun, 23 May 2010 10:11:59 -0700 (PDT), amzoti <amzoti(a)gmail.com>
wrote:

>On May 23, 6:58�am, David C. Ullrich <ullr...(a)math.okstate.edu> wrote:
>> On Sun, 23 May 2010 05:34:51 -0700 (PDT), amarj japp
>>
>>
>>
>> <amarj.ro...(a)gmail.com> wrote:
>> >Q. How many 6 digit number without repetition of digits are there such
>> >that the
>> > � � digits are all non-zero and 1 and 2 do not appear consecutively
>> >in either order.
>>
>> >Soln: The Book considers for cases
>> > � � � � (i) � Neither 1 nor 2 appears as digits : �(P 7 6) = 7!
>> > � � � � (ii) �1, but not 2 appears as a digit �: 6 (P 7 5)
>> > � � � �(iii) �2, but not 1 appears � � � � � � � �: 6 (P 7 5)
>> > � � � �(iv) Both 1 and 2 appears � � � � � � �: 2.7.4.P(6,3) +
>> >4.7.6.3.P(5,2)
>>
>> >We add them to get the total of 52080
>>
>> >My question is , if i tries to do this using indirect counting i.e
>> >Number of ways , when the 1 and 2 appears together is 2*P(8,6)
>> >If is subtract above from P(9 6) i.e P(9,6) - 2*P(8,6) = 20160
>>
>> >I don't understand how i am not getting same answer , please suggest
>> >what i am doing wrong.
>>
>> Why do you think that the number of permutations where a 1 and a 2
>> appear together is 2*P(8,6)?
>>
>> I verified the answer using brute-force calculation.
>> Extremely bad, maximally inefficient Python that
>> does the job:
>>
>> def bad(x,y):
>> � return (x,y) == (0,1) or (x,y) == (1,0)
>>
>> def ok(s):
>> � for j in range(5):
>> � �if bad(s[j],s[j+1]): return 0
>> � return 1
>>
>> ss = [()]
>> for places in range(6):
>> � ss = [s+(j,) for s in ss for j in range(9) if not j in s]
>>
>> count = 0
>> for s in ss:
>> � if ok(s):
>> � � count = count + 1
>>
>> print count #5280
>>
>> >Thanks
>>
>>
>
>Did you miss a zero or something is amiss? (52080)

That was a typo on my part - it actually returns 52080.



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