## Permutation and Combination Quiz-6

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

Q1. The number of ways of distributing 8 identical balls in 3 distinct boxes, so that none of the boxes is empty, is
•  5
•  21
•  38
•  8C3
Solution
(b) The required number of ways = 8-1C3-1=7!/2!5!=21

Q2.How many numbers divisible by 5 and lying between 3000 and 4000 that can be formed from the digits 1, 2, 3, 4, 5, 6 (repetition of digits is not allowed)?
•  6P2
•  5P2
•  4P2
•  6P3
Solution
(c) In the number which is divisible by 5 and lying between 3000 and 4000, 3 must be at thousand place and 5 must be at unit place.
Therefore rest of the digits (1, 2, 3, 4, 6) fill in two places. The number of ways= 4 P2

Q3.  The total number of ways of arranging the letters AAAA BBB CC D E F in a row such that letters C are separated from one another is
•   2772000
•  1386000
•  4158000
•  None of these
Solution
(b) We have 12 letters including 2 C's. Let us ignore 2 C's and thus we have 10 letters (4 A' s,3 B' s,1 D,1 E,1 F) and these 10 letters can be arrange in (10 !)/(4 !3 !) ways. Now, after arranging these 10 letters there will be 11 gaps in which two different letters can be arranged in 11 P2 ways. But, since 2 C's are alike, the number of arrangements will be 1/(2 !) 11 P2=(11 !)/(9 ! 2 !) So, total number of ways in which C's are separated from one another =(10 !)/(4 ! 3 !)∙(11 !)/(9 ! 2 !)=1386000

Q4. If 12 Pr=1320, then r is equal to
•  5
•  4
•  3
•  2
Solution
(c) ∵ 12Pr=1320=12×11×10
⇒ 12!/(12-r)!=12×11×10
∴ r=3

Q5. The exponent of 3 in 100 !, is
•  210
•  10!
•  1023
•  100
Solution
(c) The hall can be illuminated by switched on at least one of the 10 bulbs.
Therefore, the required number of ways is 210-1=1023

Q6. The number of ways in which 6 rings can be worn on four fingers of one hand, is
•  46
•  6C4
• 64
•  24
Solution
(a) There are 6 rings and 4 fingers. Since, each ring can be worn on any finger.
∴ Required number of ways=46

Q7. The total number of different combinations of letters which can be made from the letters of the word MISSISSIPPI is
•  150
•  148
•  149
•  None of these
Solution
(c) Here, we have 1 M,4 I' s,4 S's and 2 P's
∴ Total number of selections =(1+1)(4+1)(2+1)-1=149

Q8. 20 persons are invited for a party. In how many different ways can they and the host be seated at circular table, if the two particular persons are to be seated on either side of the host?
•  20!
•  2!×18!
•  18!
•  None of these
Solution
(b) There are total 20+1=21 persons. The two particular persons and the host be taken as one unit so that these remaining 21-3+1=19 persons be arranged in round table in 18! ways.
But the two persons on either side of the host can themselves be arranged in 2! ways
∴ required number of ways=2!×18!

Q9. 4 buses runs between Bhopal and Gwalior. If a man goes from Gwalior to Bhopal by a bus and comes back to Gwalior by another bus, then the total possible ways are
•  12
•  16
•  4
•  8
Solution
(a) Man goes from Gwalior to Bhopal in 4 ways and they come back in 3 ways.
∴ Total number of ways=4×3=12 ways

Q10. The number of ways to arrange the letters of the word CHEESE are
•  120
•  240
•  720
• 6
Solution
(a) 120

#### Written by: AUTHORNAME

AUTHORDESCRIPTION

## Want to know more

Please fill in the details below:

## Latest NEET Articles\$type=three\$c=3\$author=hide\$comment=hide\$rm=hide\$date=hide\$snippet=hide

Name

ltr
item
BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET & IIT JEE COACHING: Permutation-and-combination-quiz-6
Permutation-and-combination-quiz-6