# 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

### The Number of times the digit 8 will be written when listing the integers from 1 to 1000 is :

 A) 100 B) 200 C) 300 D) 400

Explanation:

Since 8 does not occur in 1000, we have to count the number of times 8 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, where $\inline&space;0\leq&space;x,&space;y,z\leq&space;9.$

Let us first count the numbers in which 8 occurs exactly once.

Since 8 can occur atone place in $\inline&space;^{3}\textrm{C}_{1}$ ways. There are $\inline&space;^{3}\textrm{C}_{1}(9\times&space;9)=3\times&space;9^{2}$ such numbers.

Next, 8 can occur in exactly two places in $\inline&space;^{3}\textrm{C}_{2}(9)=3\times&space;9$ such numbers. Lastly, 8 can occur in all three digits in one number only.

Hence, the number of times 8 occur is

$\inline&space;1\times&space;(3\times&space;9^{2})+2\times&space;(3\times&space;9)+3\times&space;1=300$

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

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### If the letters of the word VERMA are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word VERMA is :

 A) 108 B) 117 C) 810 D) 180

Explanation:

The number of words beign with A is 4!

The number of words beign with E is 4!

The number of words beign with M is 4!

The number of words beign with R is 4!

Number of words beign with VA is 3!

Words beign with VE are VEAMR

VEARM

VEMAR

VEMRA

VERAM

VERMA

$\inline&space;\therefore$  The Rank of the word VERMA = 4 x 4! + 3! + 6

= 96 + 6 + 6 =108

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

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### The number of signals that can be generated by using 5 differently coloured flags, when any number of them may be hoisted at a time is:

 A) 235 B) 253 C) 325 D) None of these

Explanation:

Required number of signals = $\inline&space;^{5}\textrm{P}_{1}+^{5}\textrm{P}_{2}+^{5}\textrm{P}_{3}+^{5}\textrm{P}_{4}+^{5}\textrm{P}_{5}$

= 5 + 20 + 60 + 120 + 120 = 325

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

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### In how many ways can 100 soldiers be divided into 4 squads of 10,20, 30, 40 respectively?

 A) 1700 B) 18! C) 190 D) None of these

Explanation:

$\inline&space;^{100}\textrm{C}_{10}\times&space;^{90}\textrm{C}_{20}\times&space;^{70}\textrm{C}_{30}\times&space;^{40}\textrm{C}_{40}=\frac{100!}{10!\times&space;20!\times&space;30!\times&space;40!}$

Subject: Permutation and Combination - Quantitative Aptitude - Arithmetic Ability

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### What is the total no of ways of selecting atleast one item from each of the two sets containing 6 different items each?

 A) 2856 B) 3969 C) 480 D) None of these

Explanation:

We can select atleast one item from 6 different items = $\inline&space;(2^{6}-1)$

Similarly we can select atleast one item from other set of 6 different items in $\inline&space;(2^{6}-1)$ ways.

$\therefore$ Required number of ways = $\inline&space;(2^{6}-1)(2^{6}-1)$

= $\inline&space;(2^{6}-1)^2$ = 3969