Suppose if Ajay and Vijay from the earlier example did not get along and could not picked together. In this case, we can consider three cases - one when Ajay is picked, one when Vijay is picked and one when neither is picked. The sum of these three cases is equal to the number of cases without condition - cases where they are picked together. Number of ways they can be picked together is 715 ways as seen earlier. Hence, number of ways in which at max only one of them is picked = 1365 - 715= 650 ways.
Formula
Selection without arrangement: Choosing r out of n options can be represented as
Tips
If the arrangement should always begin or end with a particular item, its order is fixed. Hence the remaining n-1 items can be arranged in (n-1)! ways.
Solved Example
MULTIPLE CHOICE QUESTION
Try yourself: How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 and 9 which are divisible by 5 and none of its digit is repeated?
A
12
B
16
C
20
D
24
E
None of these
Correct Answer: C
Answer – c) 20 Explanation : _ _ 5 first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number = 5*4*1 = 20
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MULTIPLE CHOICE QUESTION
Try yourself: In how many different ways can the letter of the word ELEPHANT be arranged so that vowels always occur together?
A
2060
B
2160
C
2260
D
2360
E
None of these
Correct Answer: B
Answer – b) 2160 Explanation : Vowels = E, E and A. They can be arranged in 3!/2! Ways so total ways = 6!*(3!/2!) = 2160
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MULTIPLE CHOICE QUESTION
Try yourself: A fruit basket contains 4 oranges, 5 apples and 6 mangoes. The number of ways a person make selection of fruits from among the fruits in the basket is?
A
269
B
280
C
209
D
256
E
None of these
Correct Answer: C
Whenever it is not explicity mentioned that fruit and distint, We take them as identical. 0 or more orange can be selected from 4 identical oranges in (4+1)=5 ways. 0 or more apples can be selected from 5 identical apples in (5+1)=6 ways.
0 or more mangoes can be selected from 6 identical mangoes in 7 ways.
∴ Total no. of ways in which all of three types of fruits can be selected (the no. of any type of fruits may also be 0),
5×6×7=210,
But in these 20 selection, there is one selection where all fruits are 0, hence we reduce 1 selection.
∴ the required no. =210−1=209
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MULTIPLE CHOICE QUESTION
Try yourself: There are 15 points in a plane out of which 6 are collinear. Find the number of lines that can be formed from 15 points.
A
105
B
90
C
91
D
95
E
None of these
Correct Answer: C
Answer – c) 91 Explanation : From 15 points number of lines formed = 15c2 6 points are collinear, number of lines formed by these = 6c2 So total lines = 15c2 – 6c2 + 1 = 91
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MULTIPLE CHOICE QUESTION
Try yourself: In how many ways 4 Indians, 5 Africans and 7 Japanese be seated in a row so that all person of same nationality sits together
A
4! 5! 7! 3!
B
4! 5! 7! 5!
C
4! 6! 7! 3!
D
can’t be determined
E
None of these
Correct Answer: A
Answer – a) 4! 5! 7! 3! Explanation : 4 Indians can be seated together in 4! Ways, similarly for Africans and Japanese in 5! and 7! respectively. So total ways = 4! 5! 7! 3!
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