There are 5 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next two have 6 choices each?

Correct Answer: 215

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.

 

Maximum number of unsuccessful attempts = 216 - 1 = 215.

Correct Answer: 11760

Explanation:

Required number of ways =  8 C 5 * 10 C 6 =   8 C 3 * 10 C 4 = 11760

Correct Answer: 2520

Explanation:

The number of ways of arranging n beads in a necklace is  ( n - 1 ) ! 2 = ( 8 - 1 ) ! 2 = 7 ! 2 = 2520 

(since n = 8)

Correct Answer: 315

Explanation:

Correct Answer: 25

Explanation:

The toys are different; The boxes are identical 

 

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways 

a. 2, 2, 1 

b. 3, 1, 1 

 

Case a. Number of ways of achieving the first option 2 - 2 - 1 

 

Two toys out of the 5 can be selected in  5 C 2 ways. Another 2 out of the remaining 3 can be selected in  3 C 2 ways and the last toy can be selected in  1 C 1 way. 

 

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2 

 

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways 5 C 2 * 3 C 2= 15 ways

 

 

Case b. Number of ways of achieving the second option 3 - 1 - 1

 

Three toys out of the 5 can be selected in  5 C 3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

 

Therefore, total number of ways of getting the 3 - 1 - 1 option is  5 C 3 = 10 = 10 ways.

 

 

 

Total ways in which the 5 toys can be packed in 3 identical boxes

 

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

Correct Answer: 4! x 5!

Explanation:

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants

As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.

The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.

Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.

Hence, the total number of ways = 4! × 5!

Correct Answer: 36

Explanation:

Required Number of ways = 4 C 2 × 3 C 2 × 2 C 1 = 36 = 36

Correct Answer: 11

Explanation:

Let the number of sides be n. 

The number of diagonals is given by ⁿC₂ - n  

^{Therefore, n}C₂ - n = 44, n>0 

ⁿC₂ - n = 44

n² - 3n - 88 = 0 

n² -11n + 8n - 88 = 0  

 

As n>0, n will not be -8. Therefore, n=11.

Correct Answer: 192

Explanation:

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

 

The number can be 4 digited but greater than 4000 or 5 digited.

 

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x  P 3 4 = 3 x 24 = 72. 

 

5 digited numbers =  P 5 5 = 5! = 120 

So the total numbers greater than 4000 = 72 + 120 = 192

Correct Answer: 63

Explanation:

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.