# 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, $\inline&space;{\color{Blue}&space;50C_{3}}$

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• Related Questions

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

$\inline x\geq 5,y\geq 5,z\geq 5 \: and \: x+y+z=21$

$\inline Let\: u\geq 0,v\geq 0,w\geq 0, then$

$\inline \therefore$ x + y + z =21

$\inline \Rightarrow$ u + 5 + v + 5 + w + 5 = 21

$\inline \Rightarrow$ u + v + w = 6

$\inline \therefore$ Total number of solutions = $\inline ^{6+3-1}\textrm{C} _{3-1} =^{8}\textrm{C} _{2}=28$

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

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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

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

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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 $\inline ^{37}\textrm{C}_{2}$. But 13 straight lines out of the given 37 straight lines pass through the same point A. Therefore instead of getting $\inline ^{13}\textrm{C}_{2}$ 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 $\inline ^{11}\textrm{C}_{2}$ points, we get only one point B.

Hence the number of intersection points of the lines is $\inline ^{37}\textrm{C}_{2}-^{13}\textrm{C}_{2}-^{11}\textrm{C}_{2}+2$ = 535

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

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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 = $\inline 4C_{1}\times 6C_{4}$ = 4 x 15 = 60

In case II the number of ways = $\inline 4C_{2}\times 6C_{3}$ = 6 x 20 = 120

In case III the number of ways = $\inline 4C_{3}\times 6C_{2}$ = 4 x 15 = 60

In case IV the number of ways = $\inline 4C_{4}\times 6C_{1}$ = 1 x 6 = 6

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

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

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In a G - 20 meeting there were total 20 people representing their own country. All the representative sat around a circular table. Find the number of ways in which we can arrange them around a circular table so that there is exactly one person between two representatives namely Manmohan and Musharraf.

 A) 2 x (17!) B) 2 x (18!) C) (3!) x (18!) D) (17!)
$\inline \therefore$ Required number of permutations = 18 x (17!) x 2 = 2 x 18!