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New answer posted
6 months agoContributor-Level 10
20. i. The permutation of 6 letters in MONDAY taken 4 at a time without repetition is
ii. The permutation of 6 letters in MONDAY when all letters are taken at a time is
6P6 = = = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
iii. The permutation of having one of the two vowels (O, A) as first letter from the word MONDAY when all letters are taken at a time is
2P1 = = 2
After fixing one of the vowel as first letter we can rearrange the remaining 5 letters taking 5 at a time
5P5 = = = 5! = 5 * 4 * 3 * 2 * 1 = 120
Therefore, total permutation when all letters are used but first letter is vowel from the word MONDAY = 2
New question posted
6 months agoNew question posted
6 months agoNew answer posted
6 months agoContributor-Level 10
19. Since no letter is repeated in the word EQUATION.
The permutation of 8 letters taken all at a time
= 8P8
=
=
= 8! [since, 0! = 1]
= 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= 40320
New question posted
6 months ago
New answer posted
6 months agoContributor-Level 10
15. The permutation of 5 different digits namely 1, 2, 3, 4, 5 taken 4 at a time is
5P4 = = = 5 * 4 * 3 * 2 * 1 = 120
The permutation of having 2 or 4 at ones place is
2P1 = = = 1 * 2 = 2
After fixing one of the even number at last digit we can rearrange the remaining four digits taking 3 at a time. i.e.
4P3 = = = 4 * 3 * 2 * 1 = 24
Therefore, total permutation of 4 digit even number using 1, 2, 3, 4, 5
= 24 * 2
= 48
New answer posted
6 months agoContributor-Level 10
14. The permutation of having even number at the last digit from the given 6 different digits namely 1, 2, 3, 4, 5, 6 to form a 3-digit number is
After taking one of the even number as last digit we can rearrange the remaining 5 digits taking 2 at a time. i.e.
Therefore, The required number = 20 * 3 = 60
New answer posted
6 months agoContributor-Level 10
49. Here, out of 100 students, first section has 40 students and the rest I e, 60 students enters in second section.
As me and my friend are among the 100 students.
The no. of ways of selecting 2 students from the 100 students
= 100C2
(a) When both enters first section if 2 of us are among the 40 students that are to be selected. Similarly, if both enters second section among the 60 students for that section.
(if 2 of us)
Hence, no. of ways of selecting both in same section = 40C2 + 60C2
Probability that both of us are in same section
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