Permutations and Combinations

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New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Let xyz be the three digit number
x+y+z=10, x≥1, y≥0, z≥0
x-1=t => x=1+t, x-1≥0
t≥0
t+y+z = 10-1
t+y+z=9, 0≤t, y, z≤9
total number of non negative integral solution =? ³? ¹C? = ¹¹C? = (11*10)/2 = 55
But for t=9, x=10, so required number or integers = 55-1=54

New question posted

a month ago

0 Follower 17 Views

New answer posted

a month ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

MOTHER
1? E
2? H
3? M
4? O
5? R
6? T
So position of word MOTHER in dictionary
2*5!+2*4!+3*3!+2!+1
=240+48+18+2+1
=309

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Family of planes: x+y+z+1+? (2x-y+z+3)=0. Parallel to line means dot product of normals is zero.

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

| Class | No. of students | Number of possible cases |
| :-: | :-: | :- |
| 10 | 5 | (I) 2 | (II) 3 | (III) 2 |
| 11 | 6 | (I) 2 | (II) 2 | (III) 3 |
| 12 | 8 | (I) 6 | (II) 5 | (III) 5 |
Total cases =? C? *? C? *? C? +? C? *? C? *? C? +? C? *? C? *? C?
= 23,800 = 100K
∴ K = 238

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