Classical

Combinatories - Combinatories MCQ

1:  

 Ramesh has 6 friends. In how many ways can he invite one or more of them at a dinner ?

A. 61
B. 62
C. 63
D. 64
 
 

Option: C

Explanation :

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Acme said: (8:16pm on Friday 11th January 2013)
Would u pls explain?
Aditya said: (10:24am on Tuesday 29th January 2013)
You seem to be looking for Sum[C[6,n], n=1..6], where C[6,n] = 6!/(n!(6-n!)) is the number of combinations of 6 things taken n at a time.c[6,1]=6!/(1!5!)=6c[6,2]=6!/(2!4!)=15c[6,3]=6!/(3!3!)=20c[6,4]=6!/(4!2!)=15c[6,5]=6!/(5!1!)=6c[6,6]=6!/(6!0!)=1sum[6,15,20,15,6,1]=63

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2:  

The number of ways to arrange th a letters of the  word CHEESE are

A. 120
B. 240
C. 720
D. 6
 
 

Option: A

Explanation :

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Sreenivasulu.G said: (12:49pm on Friday 7th April 2017)
6!/3! = 6×5×4 = 120

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3:  

The number of different permutations of the word BANANA is

A. 720
B. 60
C. 120
D. 360
 
 

Option: B

Explanation :

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4:  

The number n of ways that an organization consisting of twenty -six members can elect a president, treasurer, and secretary (assuming no person is elected to more than one position) is

A. 15600
B. 15400
C. 15200
D. 15000
 
 

Option: A

Explanation :
The president can be Plected in twenty-six
different ways; following this, the treasurer can
he elected in twenty-five different ways (since the person chosen president is :-:ot eligible to be treasurer); and following this, the secretary can be elected in twenty-four different ways.
Thus, by above principle of counting, there are n = 26 x 25 x 24 = 15 600 different ways in which the organization can elect the officers.

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5:  

Common Data for next 2 questions

There are four bus lines between A and B; and three bus lines between B and C.

The number of ways a person can travel by bus from A to C by way of B will be

A.

10

B.

12

C.

14

D.

24

 
 

Option: B

Explanation :

There are four ways to go from A to B and three ways to go from B to C.
Hence there are 4 x 3 = 12 ways to go from A to C by way of B;

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