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# If 6 different numbers are to be selected from integers 0 to

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If 6 different numbers are to be selected from integers 0 to [#permalink]

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13 Nov 2012, 20:41
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If 6 different numbers are to be selected from integers 0 to 6, how many 6-digit even integers greater than 300,000 can be composed?

Anybody with a slot method?

--- Ans will be provided later.

Thanks,
BC

Last edited by Bunuel on 14 Nov 2012, 02:01, edited 1 time in total.
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Re: Tough Probability! [#permalink]

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13 Nov 2012, 20:57
bellcurve wrote:
If 6 different numbers are to be selected from integers 0 to 6, how many 6-digit even integers greater than 300,000 can be composed?

Anybody with a slot method?

--- Ans will be provided later.

Thanks,
BC

Why not..

Since the number must be greater than 300,000 :

first approach:
Take first number 3 or more than 3. Last number 0,2,4,6. For rest anything is fine
= 4*7*7*7*7*4
But this includes 300,000 as well. Since we want greater than 300,000, substract this 1 possiblity.

Ans = 4*7*7*7*7*4 -1

Second approach:
Take first number 3 or more than 3. For rest anything is fine
= 4*7*7*7*7*7
But we want only even numbers. Last digit can have 7 possible values out of which 4 are even.
Therefore multiply above number with 4/7
Total number of intergers equal to or more than 300000 = = 4*7*7*7*7*7 *4/7

But this includes 300,000 as well. Since we want greater than 300,000, substract this 1 possiblity.

Ans = 4*7*7*7*7*7*4/7 -1

Ans = 4*7*7*7*7*4 -1

Both approaches are slot method.

Hope it helps!

Edit: Sorry dint notice. we can use only digits 0-6 not 0-9. updated.
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Last edited by Vips0000 on 13 Nov 2012, 21:38, edited 2 times in total.
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Re: Tough Probability! [#permalink]

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13 Nov 2012, 21:15
since we need to select from 7 nos. i.e. 0, 1, 2, 3 ,4, 5 and 6 and find the nos. greater than 300,000.

suppose the no. to be 123456 and we are finding nos. greater than or equal to 300,000

@position 1 - 3, 4, 5 and 6 ( we can put 4 nos.)
@positions2, 3, 4, 5 and 6 - 0, 1, 2, 3, 4, 5 and 6 (we can put 7 nos.)

(we need to deduct 1 from the answer as the question says "find the nos. GREATER than 300,000)

Therefore, the answer is (4x7x7x7x7x7) - 1 = 67,227

Hope this helps!
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Re: Tough Probability! [#permalink]

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13 Nov 2012, 23:01
bellcurve wrote:
If 6 different numbers are to be selected from integers 0 to 6, how many 6-digit even integers greater than 300,000 can be composed?

Anybody with a slot method?

--- Ans will be provided later.

Thanks,
BC

No. of ways of choosing first digit = 4 (Greater than 2, Lesser than 7)
No. of ways of selecting last digit = 4 (Even numbers Greater than 2, Lesser than 7)
No. of ways of selecting remaining four digits = 7*7*7*7 (Lesser than 7)

So answer is 7*7*7*7*4*4 - 1 = 38415

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Re: Tough Probability! [#permalink]

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13 Nov 2012, 23:59
sorry i dint see even numbers.
answer would be (4x7x7x7x7x4)-1 = 38416-1 = 38415
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Re: If 6 different numbers are to be selected from integers 0 to [#permalink]

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14 Nov 2012, 07:17
Please note the "6 different numbers" in the question; numbers cannot be repeated. If you select 4 as the first integer you can not select 4 as well as the last integer. In another words if you select even as the first number, you will only have 3 options for the last number, but if you select odd number as the first number than you will have 4 options for the last number.
Re: If 6 different numbers are to be selected from integers 0 to   [#permalink] 14 Nov 2012, 07:17
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