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If 6 coins are tossed, how many different coin sequences will have exa

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marcodonzelli
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If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   Updated on: 23 May 2015, 08:12
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If 6 coins are tossed, how many different coin sequences will have exactly 3 tails, if all tails have to occur in a row?

A. 4
B. 8
C. 16
D. 20
E. 24
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A

Originally posted by marcodonzelli on 07 Feb 2008, 12:57.
Last edited by Bunuel on 23 May 2015, 08:12, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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tizen
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If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   23 May 2015, 17:48
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You can do it using combinations. We have the restriction that we must have exactly 3 t's and they must be sequential. Whenever you have a requirement that two elements be placed next to each other, you combine them into 1 element. See we also require exactly 3 t's we know that the remaining 3 slots must by h's. Our set is now: h h ttt h.

The combinatorics question now is how many ways can we arrange 4 items, where 1 item has a repetition of 3. The formula for combinations of sets with repetitions is \(\frac{n!}{a!b!...f!}\) n is the total number of items, and a,b,...,f factorial are the repetitions for the items in the set. In our case we have h with repetition of 3 and ttt with a repetition of 1. \(\frac{4!}{3!1!}\) = 4
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   07 Feb 2008, 13:07
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C

[TTT]222 (8) - tail, tail, tail, tail or head, tail or head, tail or head
H[TTT]22 (4) - head, tail, tail, tail, tail or head, tail or head
2H[TTT]2 (4) - tail or head, head, tail, tail, tail, tail or head

N=8+4+4=16
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   07 Feb 2008, 17:21
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the question says
exactly 3 tails ..
my answer is 4 ?
am i making a mistake somewhere
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   07 Feb 2008, 20:20
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To have only three tails in consecutive order, sequence of tails should start at either 1st or 2nd or 3rd or 4th coin. No need to test all combinations because remaining coins are head.
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   08 Feb 2008, 00:42
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nikunjhd wrote:
the question says
exactly 3 tails ..
my answer is 4 ?
am i making a mistake somewhere


You are right. I was inattentive ...
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   08 Feb 2008, 14:04
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marcodonzelli wrote:
If 6 coins are tossed, how many different coin sequences will have exactly 3 tails, if all tails have to occur in a row?

4

8

16

20

24


TTTHHH
HTTTHH
HHTTTH
HHHTTT

(A)
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   06 Jul 2015, 21:40
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Exactly 3 tails and have to occur in row

We have 6 coins we use the glue method so we say all 3 tails as 1

so we have 4! ways, however we need to reduce duplicate for 3 Heads as well

HHH{T} therefore using anagram method

4!/3! = 4
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   01 Feb 2020, 09:43
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marcodonzelli wrote:
If 6 coins are tossed, how many different coin sequences will have exactly 3 tails, if all tails have to occur in a row?

A. 4
B. 8
C. 16
D. 20
E. 24



All 3 tails have to occur in a row. We can, thus, count the 3 tails as a single item.
This leaves us with 3 coins that show heads. However, the heads can occur in any position, not necessarily in a row.

Thus, the 3 tails (counted as 1 item) and the 3 heads make for 4 items of which the 3 heads are identical; as shown below:

(T T T), H, H, H

Thus, total number of arrangements = \(\frac{4!}{3!} = 4\)

Answer A
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If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   10 Nov 2023, 22:49
Hi Bunuel,
Can you respond to this question ,it seems ambiguous. The question is unclear whether the remaining 3 outcomes are "head" or "can be Head or Tail".
In case 1, The solution would be 4!/3!=4
In case 2, The solution would be 4!/3!+4!/2!+4!/2!+4!/3!=30 (I have taken 3H 0T and Sequence + 2H 1T and sequence +1H 2T and sequence +0H 3T and sequence , where sequence is TTT.

Correct me if I am wrong.

Thanks
Umang
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   11 Nov 2023, 01:55
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umangshah wrote:
Hi Bunuel,
Can you respond to this question ,it seems ambiguous. The question is unclear whether the remaining 3 outcomes are "head" or "can be Head or Tail".
In case 1, The solution would be 4!/3!=4
In case 2, The solution would be 4!/3!+4!/2!+4!/2!+4!/3!=30 (I have taken 3H 0T and Sequence + 2H 1T and sequence +1H 2T and sequence +0H 3T and sequence , where sequence is TTT.

Correct me if I am wrong.

Thanks
Umang


Have you missed the word "exactly" in the stem?

    If 6 coins are tossed, how many different coin sequences will have exactly 3 tails, if all tails have to occur in a row?
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink] New post   11 Nov 2023, 04:01
Bunuel wrote:
umangshah wrote:
Hi Bunuel,
Can you respond to this question ,it seems ambiguous. The question is unclear whether the remaining 3 outcomes are "head" or "can be Head or Tail".
In case 1, The solution would be 4!/3!=4
In case 2, The solution would be 4!/3!+4!/2!+4!/2!+4!/3!=30 (I have taken 3H 0T and Sequence + 2H 1T and sequence +1H 2T and sequence +0H 3T and sequence , where sequence is TTT.

Correct me if I am wrong.

Thanks
Umang


Have you missed the word "exactly" in the stem?

    If 6 coins are tossed, how many different coin sequences will have exactly 3 tails, if all tails have to occur in a row?


I thought that exactly refers only for sequence of Tails. Now I got it.. Thanks :)
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Re: If 6 coins are tossed, how many different coin sequences will have exa [#permalink]
11 Nov 2023, 04:01
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