#### Permutations and Combinations 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: Permutations and Combinations - 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

Using numbers from 0 to 9 the number of 5 digit telephone numbers that can be formed is

A) 1,00,000 | B) 59,049 |

C) 3439 | D) 6561 |

Explanation:

The numbers 0,1,2,3,4,5,6,7,8,9 are 10 in number while preparing telephone numbers any number can be used any number of times.

This can be done in ways, but '0' is there

So, the numbers starting with '0' are to be excluded which one in number.

Total 5 digit telephone numbers = = 3439

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Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

In how many ways the letters of the word 'DESIGN' can be arranged so that no consonant appears at either of the two ends?

A) 240 | B) 72 |

C) 48 | D) 36 |

Explanation:

DESIGN = 6 letters

No consonants appears at either of the two ends.

=

= 2 x 4 x 3 x 2 x 1

= 48

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Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

In How many ways can the letters of the word 'CAPITAL' be arranged in such a way that all the vowels always come together?

A) 360 | B) 720 |

C) 120 | D) 840 |

Explanation:

CAPITAL = 7

Vowels = 3 (A, I, A)

Consonants = (C, P, T, L)

5 letters which can be arranged in

Vowels A,I =

No.of arrangements = =360

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Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

The number of ways that 8 beads of different colours be strung as a necklace is

A) 2520 | B) 2880 |

C) 4320 | D) 5040 |

Explanation:

The number of ways of arranging n beads in a necklace is (since n = 8)

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Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability

The number of ways that 7 teachers and 6 students can sit around a table so that no two students are together is

A) 7! x 7! | B) 7! x 6! |

C) 6! x 6! | D) 7! x 5! |

Explanation:

The students should sit in between two teachers. There are 7 gaps in betweeen teachers when they sit in a round table. This can be done in ways. 7 teachers can sit in (7-1)! ways.

Required no.of ways in =

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Subject: Permutations and Combinations - Quantitative Aptitude - Arithmetic Ability