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Bunuel
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If 5 integers are to be selected at random and without replacement 18 Apr 2026, 00:08
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Isha2701
Bunuel - what does without replacement mean here? I understand that we choose 10 and 20 def, meaning 1/20 x 1/19 x 18/18 x 17/17 x 16/16 (then for rest three numbers bcause we have 18 left now and any 18 can be chosen, similarly for 17 and 16 left.) And they all can be arranged in 5! ways..so where am I wrong?
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“Without replacement” means that once a number is chosen, it is removed from the pool, so it cannot be chosen again and the next denominator drops from 20 to 19, then 18, and so on. Your mistake is using 5!: the three “any” positions are not distinct, so swapping those among themselves does not create new cases, which is why you count only 5!/3! = 20 arrangements, not 120.

For more please review the rest of the discussion.
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If 5 integers are to be selected at random and without replacement 14 Jul 2026, 00:28
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Prob = (2C2 * 18C3)/20C5 = 1/19
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Elmon
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If 5 integers are to be selected at random and without replacement 14 Jul 2026, 07:32
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Remind me:

even though we switch the order 10 & 20, we don't need to multiple 2 times.
Bunuel
­If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

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If 5 integers are to be selected at random and without replacement 15 Jul 2026, 08:56
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Hi Elmon,

Yes, you've got it right, and here's the clean reason to lock it in.

This is exactly the point Bunuel made earlier in the thread when he shot down the 2/20 * 1/19 approach: multiplying by 2 ("10 first or 20 first") double-counts the same selection.

Think about what the question actually asks: it wants both 10 and 20 to be among your 5 picks. It does not care whether 10 landed in your hand before 20 or after. The set {10, 20, ...} is one and the same outcome either way.

- "Draw 10, then 20" - you end up holding {10, 20}
- "Draw 20, then 10" - you end up holding {10, 20}

Same result. So if you multiply by 2, you're counting that one favorable outcome twice - which is why the answer would come out wrong.

That's the whole spirit of the combination method everyone used: 18C3 / 20C5. Combinations already ignore order, so there's simply no order to "restore" with a factor of 2.

Quick way to feel it

Shrink the problem. Suppose you just pick 2 numbers from the tiny set {10, 20}. List every distinct selection by hand:

- {10, 20}

That's it - one outcome, not two. You never write down {10, 20} and {20, 10} as separate picks, because a selection has no order. The moment you'd multiply by 2, you'd be pretending those are two different things.

So your instinct is correct: no factor of 2. Order only matters when the problem itself treats different arrangements as different outcomes - and choosing a group of integers isn't one of those.

Answer: D

Elmon
Remind me:

even though we switch the order 10 & 20, we don't need to multiple 2 times.

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PGTLrowanhand
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If 5 integers are to be selected at random and without replacement 06 Aug 2026, 14:31
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Here's my video solution:
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