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

Question

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

tizen · Accepted Answer

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

walker · Answer

C

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

N=8+4+4=16

nikunjhd · Answer

the question says 
exactly 3 tails ..
my answer is 4 ?
am i making a mistake somewhere

Travel09 · Answer

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.

walker · Answer

You are right. I was inattentive ...

incognito1 · Answer

TTTHHH
HTTTHH
HHTTTH
HHHTTT

(A)

kuttingchai · Answer

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

sujoykrdatta · Answer

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

umangshah · Answer

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

Bunuel · Answer

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?

umangshah · Answer

I thought that exactly refers only for sequence of Tails. Now I got it.. Thanks

Rohit_842 · Answer

Even if it were case 2. Total should've been 15 and not 30, 4!/3!+4!/2!2!+4!/3!+4!/4!=15. (I know it's not what the question is asking but I'm giving it a try just to improve my hand in counting and combinatorics as I'm really bad at it.) Am I right?  Bunuel

Krunaal · Answer

Your calculation of 4!/3!+4!/2!2!+4!/3!+4!/4!=15 is right. However, if it were case 2 (atleast 3 Tails) the answer would be 10. The correct answer for this particular question is 4.

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

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

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