# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?
 ✖ A. $2^5$ ✖ B. $41$ ✖ C. $22$ ✔ D. $42$ ✖ E. $31$

Solution:
Option(D) is correct

The question requires you to find a number of the outcomes in which at most $3$ coins turn up as heads.

i.e., $0$ coins turn heads or $1$ coin turns head or $2$ coins turn heads or $3$ coins turn heads.

The number of outcomes in which $0$ coins turn heads is ${^6C_0} = 1$ outcome

The number of outcomes in which $1$ coin turns head is ${^6C_1} = 6$ outcomes

The number of outcomes in which $2$ coins turn heads is ${^6C_2} = 15$ outcomes

The number of outcomes in which $3$ coins turn heads is ${^6C_3} = 20$ outcomes.

Therefore, total number of outcomes,

$= 1 + 6 + 15 + 20$

$= 42\text{ outcomes.}$

Edit: Corrected the typo after it was pointed by Kiran.

Edit 2: For an alternative solution, check comment by ABHIJEET.

## (5) Comment(s)

Sandeep
()

How is ${^6C_3} = 20$? Is it not 40$? Sandeep () Sorry...............got it! ABHIJEET () another way of solving the question is by arranging coins in different cases 1.0 Coins turn heads (T,T,T,T,T,T) possible arrangement = 6!/6! = 1 2. 1 coin turns H (H,T,T,T,T,T) possible arrangement = 6!/5! = 6 3. 2 coins turns H (H,H,T,T,T,T) possible arrangement = 6!/(2! x 5!) = 15 4. 3 coins turns H (H,H,H,T,T,T) possible arrangement = 6!/(3! x 3!) = 20 all possible = 1+6+15+20 = 42 ABHIJEET () 3rd case.. 6!/(2! x 4!) wrote 5! by mistake.. Kiran () The number of outcomes in which 3 coins turn heads is$^6C_0=20$outcomes. where$^6C_0$is$^6C_3\$

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