Aptitude 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.**

**ABHIJEET**

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3rd case.. 6!/(2! x 4!) wrote 5! by mistake..

**Kiran**

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The number of outcomes in which 3 coins turn heads is $^6C_0=20$ outcomes.

where $^6C_0$ is $^6C_3

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