#### Permutation and Combination Question & Answers

From a deck of 52 cards, a 5 card hand is dealt.How may distinct five card hands are there if the queen of spades and the four of diamonds must be in the hand?

A) 52C5 | B) 50C3 |

C) 52C4 | D) 50C4 |

Explanation:

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.

So,

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

Tags: From a deck of 52 cards a 5 card hand is dealt.How may distinct five card hands are there if the queen of spades and the four of diamonds must be in the hand

- Related Questions

Find the value of 'n' for which the nth term of two AP'S:

15,12,9.... and -15,-13,-11...... are equal?

A) n = 2 | B) n = 5 |

C) n = 29/5 | D) n = 1 |

Explanation:

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1)

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5

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Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic AbilityJob Role: Analyst

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

A) 28 | B) 14 |

C) 21 | D) 7 |

Explanation:

Let x, y, z be the number of balls received by the three persons, then

x + y + z =21

u + 5 + v + 5 + w + 5 = 21

u + v + w = 6

Total number of solutions =

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Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

In How many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?

A) 131 | B) 191 |

C) 68 | D) 3720 |

Explanation:

Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind.

Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191

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Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines.

A) 525 | B) 535 |

C) 545 | D) 555 |

Explanation:

The number of points of intersection of 37 lines is . But 13 straight lines out of the given 37 straight lines pass through the same point A. Therefore instead of getting points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting points, we get only one point B.

Hence the number of intersection points of the lines is = 535

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Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady?

A) 123 | B) 113 |

C) 246 | D) 945 |

Explanation:

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.

(i) 1 lady out of 4 and 4 gentlemen out of 6

(ii) 2 ladies out of 4 and 3 gentlemen out of 6

(iii) 3 ladies out of 4 and 2 gentlemen out of 6

(iv) 4 ladies out of 4 and 1 gentlemen out of 6

In case I the number of ways = = 4 x 15 = 60

In case II the number of ways = = 6 x 20 = 120

In case III the number of ways = = 4 x 15 = 60

In case IV the number of ways = = 1 x 6 = 6

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

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Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability