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# How many positive integers of four different digits, each

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Intern
Joined: 21 Aug 2010
Posts: 5
How many positive integers of four different digits, each  [#permalink]

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12 Sep 2010, 07:47
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1) How many positive integers of four different digits, each greater than 3000, can be formed with the digits 1, 2, 3, 4, 5 and 6?

2) How many positive integers of three different digits, each less than 400 can be formed from the digits 1, 2, 3, 4, 5 and 6?
Manager
Joined: 20 Jul 2010
Posts: 213
Re: PS - P & C  [#permalink]

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12 Sep 2010, 08:09
A)
Number of digits that can take different place
3 5 4 3
_ _ _ _

3*5*4*3 = 180 numbers

B

3 5 4
_ _ _
Number below 400 = 3*5*4 = 60
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Joined: 02 Sep 2009
Posts: 51184
Re: PS - P & C  [#permalink]

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12 Sep 2010, 09:59
jinx83 wrote:
1) How many positive integers of four different digits, each greater than 3000, can be formed with the digits 1,2,3,4,5 and 6?

2) How many positive integers of three different digits, each less than 400 can be formed from the digits 1,2,3,4,5 and 6?

Post one question per topic and provide answer choices.

One can do the way saxenashobhit proposed or:

1. How many positive integers of four different digits, each greater than 3000, can be formed with the digits 1, 2, 3, 4, 5 and 6?

# of 4-digit numbers possible to form with 6 digits 1, 2, 3, 4, 5, and 6 is $$P^4_6=360$$. As number must be more than 3,000 then it should start with 3, 4, 5, or 6 (with 4 out of 6 possible) --> so every 4 out of 6 number from 360 will fit --> $$\frac{4}{6}*360=240$$.

2. How many positive integers of three different digits, each less than 400 can be formed from the digits 1, 2, 3, 4, 5 and 6?

# of 3-digit numbers possible to form with 6 digits 1, 2, 3, 4, 5, and 6 is $$P^3_6=120$$. As number must be less than 400 then it should start with 1, 2, or 3 (with 3 out of 6 possible) --> so every 3 out of 6 number from 240 will fit --> $$\frac{3}{6}*120=60$$.

saxenashobhit wrote:
A)
Number of digits that can take different place
3 5 4 3
_ _ _ _

3*5*4*3 = 180 numbers

B

3 5 4
_ _ _
Number below 400 = 3*5*4 = 60

One typo: you should have 4 instead of 3.

Hope it's clear.
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Re: PS - P & C  [#permalink]

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12 Sep 2010, 10:16
Bunuel wrote:
2. How many positive integers of three different digits, each less than 400 can be formed from the digits 1, 2, 3, 4, 5 and 6?

# of 3-digit numbers possible to form with 6 digits 1, 2, 3, 4, 5, and 6 is $$P^3_6=120$$. As number must be less than 400 then it should start with 1, 2, or 3 (with 3 out of 6 possible) --> so every 3 out of 6 number from 240 will fit --> [highlight]$$\frac{3}{6}*240=180$$[/highlight].

Bunuel, for the solution to the second question. Is there a typo? Should it not be $$(3/6)*120$$.
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Joined: 03 Apr 2010
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Re: PS - P & C  [#permalink]

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12 Sep 2010, 11:04
saxenashobhit wrote:
A)
Number of digits that can take different place
3 5 4 3
_ _ _ _

3*5*4*3 = 180 numbers

B

3 5 4
_ _ _
Number below 400 = 3*5*4 = 60

thanx for solving!!
Manager
Joined: 20 Jul 2010
Posts: 213
Re: PS - P & C  [#permalink]

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12 Sep 2010, 20:13
Thanks Bunuel. Silly mistake on my part. Digits 3,4,5 and 6 will result in number greater than 3000. So 4, I agree.
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Re: PS - P & C  [#permalink]

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12 Sep 2010, 23:59
1
1)

the required # is > 3000
implies, The 1000s place shud be taken by 3,4,5,6, hence 4 posibilities. The remaining 3 digits can be selected from the remaining 5 available #s (note that one digit 3/4/5/6 is already selected for 1000s digit). Hence 5C3 ways, and those 3 #s can be arranges in 3! ways.

Hence answer is 4*5C3*3! = 240

2)

For the # to be < 400 the 100s digit shud be < 4 => one out of 3 choices 1/2/3

if the 100s digit is 1 then 10s digit can be selected in 5C1 and the units digit can be selected in 4C1 ways ==> 5C1*4C1=20
the same is the case with 2 and 3 too.

hence answer is 20+20+20 = 60
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Re: How many positive integers of four different digits, each  [#permalink]

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22 Aug 2018, 14:42
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Re: How many positive integers of four different digits, each &nbs [#permalink] 22 Aug 2018, 14:42
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