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Re: John wrote a phone number on a note that was later lost
[#permalink]
28 May 2020, 22:02
abhimahna wrote:
Ashishsteag wrote:
stne wrote:
Hi
Looking for some assistance with this question , please .
XXXX(1}{1}{1} X can be a prime number or non prime number Prime number:2,3,5,7 Non prime-4,6,8,9(1 is already taken) Probability(X is a Prime number)=4/8=1/2 Probability(X is a non-prime number)=4/8=1/2
So,Atleast 2 prime numbers=P P NP NP=4!/2!2!*1/2*1/2*1/2*1/2=6/16 Three prime numbers and one Non prime=P P P NP=4!/3!*1//2^4=4/16 All prime=PPPP=1/16 Total=11/16.
Remember that its a telelphone number,so different arrangements give you different number-same goes for codes,passwords,words..
Can you please explain me why you have multiplied with 4!/2!2!? We do have 2 primes but they could be different, right?
Apologies for reviving this old post. Hope this helps others who have the same doubt. I'm not sure if this query of yours was answered. Let's consider an example, 2368. The number of ways we can arrange this is 2368, 2638, 2683, 6238, 6283, 6823. Point to be noted here is that we are maintaining the same order for (2,3) and (6,8). So we can have 4 orders possible and for each order we can have 6 such arrangements. 4*6=24=4!
To sum it up, the 4 different orders (2,3),(3,2),(6,8),(8,6) are accounted as separate when we consider the possible arrangements as [P1]x[P2]x[NP1]x[NP2]. If we consider the possible arrangement for each such combination as 4!, then we'll end up having duplicates as the same will be present when we consider [P2]x[P1]x[NP1]x[NP2], [P1]x[P2]x[NP2]x[NP1], [P2]x[P1]x[NP2]x[NP1]. Therefore, to avoid duplicate scenario, we divide 4! by 2! for each prime and non-prime possible combinations.
Note: It is not related to P1=P2 or P1!=P2 or NP1=NP2 or NP1!=NP2. It's because the 4! arrangements will have 18 such possibilities which will be duplicated in other arrangments of 4P*4P*NP4*NP4.
Thanks Lipun
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Re: John wrote a phone number on a note that was later lost [#permalink]