Permutations & Combinations – IV

Required number of ways = (¹²C₂*13) + (¹³C₂*12) = 1794

Note
We add here because the events are mutually exclusive.

A rectangle with m rows will have (m+1) horizontal lines; similarly, n columns will generate (n+1) vertical lines. A rectangle of any size in this grid is created from 2 horizontal lines and 2 vertical lines.

-> Number of rectangles will correspond to the number of ways in which 2 horizontal lines can be chosen from (m+1) horizontal lines and 2 vertical lines can be chosen from (n+1) vertical lines.

-> Number of rectangles = R = ^m+1C₂ * ⁿ⁺¹C₂ = ((m+1)*m*(m-1)!)/((m+1-2)*2!) * ((n+1)*n*(n-1)!)/((n+1-2)*2!) = (m(m+1)*n(n+1))/4

The given grid is a rectangle of size 4*5 [4 rows, 5 columns]
Now, the number of rectangles, R, in a rectangular grid of size m*n is given by:
R = (m*(m+1)*n*(n+1))/4
Here, m = 4 and n = 5
R = (4*5*5*6)/4 = 150

Mechanical can sit in (5+20+10)! = 35! ways
Civil can sit in (3+15+20)! = 38! ways
Electrical can sit in (20+5+2)! = 27! ways
These 3 groups can be themselves arranged in 3! ways
Desired number of ways = 3!*35!*38!*27!

Total seating ways = (27+35+38)! = 100!

If we consider all the Mechanical engineers as 1, then the total number of students = 27+38+1 = 66, who can be arranged in 66! ways.

Therefore, no. of ways in which the Mechanical engineers sit together will be 66!*35! (within one group, they can be arranged in 35! ways)

Therefore, number of ways in which they are not together
100! – (35!*66!)

If she invites only one friend out of 7, she can do it in ⁷C₁ ways; if she invites 2 friends, she can do it in ⁷C₂ ways and so on. If she invites all her friends, she can do it in ⁷C₇ ways.
Therefore, the total number of ways of inviting friends =
⁷C₁+⁷C₂+⁷C₃+⁷C₄+⁷C₅+⁷C₆+⁷C₇
= ⁷C₁+⁷C_2­+⁷C₃+⁷C₃+⁷C₂+⁷C₁+⁷C₀   [ⁿC_r = ⁿC_n-r]
= 7 + (7*6)/(2*1) + (7*6*5)/(3*2*1) +(7*6*5)/(3*2*1)+ (7*6)/(2*1) + 7 + 1
= 7+21+35+35+21+7+1 = 127

The candidate can fail by failing in 1 or 2 or 3 or 4 or 5 subjects.
Therefore, required number of ways in which he can fail =
⁵C₁+⁵C₂+⁵C₃+⁵C₄+⁵C₅
= ⁵C₁+⁵C₂+⁵C₂+⁵C₁+⁵C₅  [ⁿC_r = ⁿC_n-r]
= 5+ (5*4)/(2*1)+(5*4)/(2*1)+ 5+1 = 31

The various possibilities are shown in the table below:

	Man’s Relatives		Woman’s Relatives
	Indians	Foreigners	Indians	Foreigners
I.	3	0	0	3
II.	2	1	1	2
III.	1	2	2	1
IV.	0	3	3	0

Number of ways for I = ⁴C₃ * ⁴C₃ —– ❶

(3 Indians, man’s relatives, can be selected from 4 Indians, man’s relatives in ⁴C_3­ ways; 3 foreigners, woman’s relatives, can be selected from 4 foreigners, woman’s relatives in ⁴C₃ ways. Applying fundamental principle of counting*, total number of ways = ⁴C₃ * ⁴C₃)

Similarly, number of ways for II = ⁴C₂*³C₁*³C₁*⁴C₂ —– ❷
Similarly, number of ways for III = ⁴C₁*³C₂*³C₂*⁴C₁ —– ❸
Similarly, number of ways for IV = ³C₃*³C₃ —– ❹

Now, ⁿC_r = n!/(n-r)!r!
❶ = ⁴C₃ * ⁴C₃ = 4!/(3!*!) * 4!/(3!*!) = 4*4 = 16
❷ = ⁴C₂*³C₁*³C₁*⁴C₂ = 4!/(2!*2!) * 3!/(3!*1!) * 3!/(2!*1!) * 4!/(2!*2!) = (4*3)/2*3*3*(4*3)/2 = 324
❸ = ⁴C₁*³C₂*³C₂*⁴C₁ = 4!/(3!*1!) * 3!/(3!*1!) * 3!/(2!*1!) * 4!/(3!*1!) = 4*3*3*4 = 144
❹ = ³C₃*³C₃ = 1*1 = 1
Total = 16+324+144+1 = 485

Note

We add here because, ❶, ❷, ❸ and ❹ are mutually exclusive events (i.e., if one happens, the others do not happen).

6 Xs can be put in 9 boxes in ⁹C₆ ways
Now ⁹C₆ = 9!/(6!*3!) = (9*8*7*6!)/(6!*3*2) = 84
Also, there will be 6 ways in which one row or one column will be vacant.
Therefore, required number of ways = 84-6 = 78

1W + 4M = ⁴C₁*⁶C₄ = 60 ways
2W + 3M = (⁴C₂ – 1) *⁶C₃ = 100 ways
3W + 2M = (⁴C₃ – 2)*⁶C₂ = 30 ways
Total = 60+100+30 = 190 ways