4
Q:

In the below word how many words are there in which R and W are at the end positions?

RAINBOW

A) 120 B) 180
C) 210 D) 240

Answer:   D) 240



Explanation:

When R and W are the first and last letters of all the words then we can arrange them in 5!ways. Similarly When W and R are the first and last letters of the words then the remaining letters can be arrange in 5! ways.

Thus the total number of permutations = 2 x 5!  = 2 x 120 = 240

Q:

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
 
Answer & Explanation Answer: C) 300

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 0x,y,z9.

 

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

 

Since 8 can occur atone place in 3C1ways. There are 3 x 9 x 9  such numbers.

 

Next, 8 can occur in exactly two places in 3C2×9 such numbers. Lastly, 8 can occur in all three digits in one number only.

  

Hence, the number of times 8 occur is 

 1×3×92+2×3×9+3×1 = 300

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2 5142
Q:

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
 
Answer & Explanation Answer: A) 108

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 

Therefore, The Rank of the word VERMA = 4 x 4! + 3! + 6 = 96 + 6 + 6 =108             

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4 7631
Q:

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
 
Answer & Explanation Answer: C) 325

Explanation:

Required number of signals = 5P1 + 5P2 + 5P3 + 5P4 + 5P5

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

 

 

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4 5664
Q:

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
 
Answer & Explanation Answer: D) None of these

Explanation:

100C10 + 90C20 + 70C30 + 40C40 = 100!10!x20!x30!x40!

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4 9620
Q:

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
 
Answer & Explanation Answer: B) 3969

Explanation:

We can select atleast one item from 6 different items = 26-1  

 

Similarly we can select atleast one item from other set of 6 different items in 26-1 ways. 

 

Required number of ways = (26-1)26-1 = 26-12 = 3969

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1 6986
Q:

If a+b+c =21 what is the total number of positive integral solutions?

A) 109 B) 190
C) 901 D) 910
 
Answer & Explanation Answer: B) 190

Explanation:

Number of positive integral solutions =  n-1Cr-1 = C220 = 190

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6 9223
Q:

If a+b+c = 21, what is the total number of non - negative integral solutions?

A) 123 B) 253
C) 321 D) 231
 
Answer & Explanation Answer: B) 253

Explanation:

Number of non - negative integral solutions =n+r-1 Cr-1 = C223 = 253 

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4 7115
Q:

In how many different ways can 6 different balls be distributed to 4 different boxes, when each box can hold any number of ball?

A) 2048 B) 1296
C) 4096 D) 576
 
Answer & Explanation Answer: C) 4096

Explanation:

Every ball can be distributed in 4 ways. 

Hence the required number of ways = 4 x 4 x 4 x 4 x 4 x 4 = 4096

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0 7629