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# Thread: challenge problems

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## challenge problems

How many different positive integers exist between 10^6and 10^7, the sum of whose digits is equal to 2?
A. 6

B. 7
C. 5
D. 8

E. 18

In how many ways can the letters of the word “PROBLEM” be rearranged to make 7 letter words such that
none of the letters repeat?

A. 7

B. 7C7
C. 77
D. 49
E. None of these

Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

A. 210

B. 29

C. 3*28

D. 3*29

E. None of these

In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the
three post boxes?

A. 5 C 3

B. 5 P 3

C. 53

D. 35

E. 25

A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit

his target?
A. 1
B.1/256
C.81/256
D.175/256

E.108/256

There are 6 boxes numbered 1, 2 ...6. Each box is to be filled up either with a red or a green ball in such a way
that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The
total number of ways in which this can be done is
A. 5
B. 21
C. 33
D. 60
E. 6

What is the probability that the position in which the consonants appear remain unchanged when the letters of
the word Math are re-arranged?
A. 1/4
B. 1/6
C. 1/3
D. 1/24
E. 1/12

What is the probability that the position in which the consonants appear remain unchanged when the letters of
the word Math are re-arranged?

A. 1/4

B. 1/6
C. 1/3

D. 1/24
E. 1/12

How many different four letter words can be formed (the words need not be meaningful) using the letters of the
word MEDITERRANEAN such that the first letter is E and the last letter is R?

A. 59

B.11!/3!*2!*2!*2!
C. 56
D. 23
E.11!/2!*2!*2!

In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear
together?

A.6!/2!

B. 3!*3!
C.4!/2!
D.4!*3!/2!

E.3!*3!/2!

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Q. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear
together?
A.6!/2!
C.4!/2!
D.4!*3!/2!
B. 3!*3!
E.3!*3!/2!

Answer is 4!* 3!/2!
Explanation: If we take all vowel as a single letter then BCS(AAU) total 4 letter can be arranged in 4! way.

Now three vowel can be arranged in 3!/2! ways. ( As in three vowel there is 2 'A' for this it is divided by 2!)
So total way is 4!*3!/2!

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What is the probability that the position in which the consonants appear remain unchanged when the letters of the word Math are re-arranged?
A. 1/4
B. 1/6
C. 1/3
D. 1/24
E. 1/12

There are four letter in math and we can rearrange this four letter in 4! ways.
And constant remain in order word is 4C
1 (math,amth, mtah and mtha) =4
So probability is 4/4
!=4/(4*3*2*1)=1/6

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How many different positive integers exist between 10^6and 10^7, the sum of whose digits is equal to 2?
A. 6

B. 7
C. 5
D. 8
E. 18

5. Good post? |
In how many ways can the letters of the word “PROBLEM” be rearranged to make 7 letter words such that
none of the letters repeat?

A. 7

B. 7C7
C. 77
D. 49
E. None of these

Answer should be E. Because the actual result should 7!

6. Good post? |
There are 6 boxes numbered 1, 2 ...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
A. 5
B. 21
C. 33
D. 60
E. 6

Answer is B=21 (6+5+4+3+2+1)

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Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

A. 210

B. 29

C. 3*28

D. 3*29

E. None of these

The answer should 2
9. Is answer B 29 or 29????

8. Good post? |
Originally Posted by Md. Minuddin

Q. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear

together?

A.6!/2!

C.4!/2!

D.4!*3!/2!

B. 3!*3!
E.3!*3!/2!

Answer is 4!* 3!/2!

understood the 4!*3! i.e. 3! is for vowels and 4! is for all letters. y is it divided by 2!.

9. Good post? |
Originally Posted by Md. Minuddin
What is the probability that the position in which the consonants appear remain unchanged when the letters of the word Math are re-arranged?

A. 1/4

B. 1/6

C. 1/3
D. 1/24
E. 1/12

Total arrangement=4!
With consonant in the same order=4

So probability is 4!/4

plz explain. i thought the answer is 1/24.

10. Good post? |
Originally Posted by Md. Minuddin
In how many ways can the letters of the word “PROBLEM” be rearranged to make 7 letter words such that
none of the letters repeat?

A. 7
B. 7C7
C. 77
D. 49
E. None of these

Answer should be E. Because the actual result should 7!

OA-A=7. not sure why