A) If (i) nP3 = nP4 find the value of n. (ii) (3!)(P4) = ? B) A committee of 5 is to be formed from 6 doctors and 5 nurses. Find total no. of ways if committee consists of i) All d... A) If (i) nP3 = nP4 find the value of n. (ii) (3!)(P4) = ? B) A committee of 5 is to be formed from 6 doctors and 5 nurses. Find total no. of ways if committee consists of i) All doctors ii) 3 doctors 2 nurses iii) No restrictions. C) Solve graphically the L.P.P. Minimize Z = 7x + y Subjectto 5x + y ≥ 5 x + y ≥ 3 x ≥ 0, y ≥ 0 D) A toy manufacturer produces toy cars and toy planes, each of which must be processed through two machines A and B. Machine A has maximum of 120 hours available and machine B has a maximum of 180 hours available. Manufacturing a car requires 6 hours in machine A and 3 hours in machine B. Manufacturing a plane requires 4 hours in machine A and 10 hours in machine B. If the profits are rupees 45 for a car and rupees 55 for a plane, formulate the LPP which will maximize the profit. Read more

Steps to Solve

1. Finding the value of ( n )

To find the value of ( n ) such that ( P_3 = P_4 ), we can use the formula for permutations.

Using the formula for permutations, we have:

[ P_n^r = \frac{n!}{(n-r)!} ]

Set up the equation as follows:

[ P_n^3 = P_n^4 \implies \frac{n!}{(n-3)!} = \frac{n!}{(n-4)!} ]

2. Solving the equations for part A

Cancel ( n! ) from both sides:

[ (n-3)! = (n-4)! ]

This simplifies to:

[ n-3 = 1 \implies n = 4 ]

3. Combinations of doctors and nurses for part B

A committee of 5 needs to be formed from 6 doctors and 5 nurses. The combinations for each case are calculated as follows:

• Case i: All doctors

[ \text{Ways} = \binom{6}{5} = 6 ]

• Case ii: 3 doctors and 2 nurses

[ \text{Ways} = \binom{6}{3} \times \binom{5}{2} ]

Calculating those:

[ \binom{6}{3} = 20 \quad \text{and} \quad \binom{5}{2} = 10 ]

Thus,

[ \text{Ways} = 20 \times 10 = 200 ]

• Case iii: No restrictions

For 5 people from 11 total (6 doctors + 5 nurses):

[ \text{Ways} = \binom{11}{5} = 462 ]

4. Solving the Linear Programming Problem (LPP) for part C

Objective function to minimize:

[ Z = 7x + y ]

Constraints:

1. ( 5x + y \geq 5 )

2. ( x + y \geq 3 )

3. ( x \geq 0 ), ( y \geq 0 )

4. Graphical Representation

Graph the lines represented by each constraint to find the feasible region and locate the vertices. Identify which vertex minimizes ( Z ).

6. Optimizing the Profit for part D

Using the maximum hours of Machine A (180 hours) and available hours per task:

• Let ( x ) be the number of cars and ( y ) the number of planes.

The constraints and profit are set up as follows:

[ 6x + 4y \leq 180 \quad \text{(Machine A)} ] [ 3x + 10y \leq 120 \quad \text{(Machine B)} ] Maximize profit ( P = 45x + 55y ).