7people leave their bags outside temple and returning after worshiping the deity picked one bageach at random. In how many ways at least one and at most three of them get their correct bags ? (A)7C3.9 +7C5.44 +7C1.265(B)7C6 .265 + 7C5 .9 + 7C7 .44(C)7C5.9 +7C2.44 +7C1.265(D)6C2.21 + 7C5.44 + 7C6. 265
Answers
Given : 7people leave their bags outside temple and returning after worshiping the deity picked one bag each at random.
To Find : In how many ways at least one and at most three of them get their correct bags
Step-by-step explanation:
n people n bags
no one gets correct
The formula for the derangement can be used directly. The number of derangements is
n! (1/2!−1/3!+1/4!+…+(−1)ⁿ/n!)
at least one and at most three of them get their correct bags
1 gets correct 6 Does not
2 gets correct 5 Does not
3 gets correct 4 Does not
1 gets correct 6 Does not
= ⁷C₁* 6! ( 11/2! - 1/3! + 1/4! - 1/5! + 1/6!)
= ⁷C₁* ( 6!/2! - 6/3! + 6!/4! - 6!/5! + 6!/6!)
= ⁷C₁* ( 360 - 120 + 30 - 6 + 1)
= ⁷C₁* 265
2 gets correct 5 Does not
= ⁷C₂* 5! ( 1/2! - 1/3! + 1/4! - 1/5! )
= ⁷C₂* ( 5!/2! - 5/3! + 5!/4! - 5!/5! )
=⁷C₂* ( 60 - 20 + 5- 1)
= ⁷C₂* 44
⁷C₂ = ⁷C₅
= ⁷C₅* 44
3 gets correct 3 Does not
= ⁷C₃* 4! ( 1/2! - 1/3! + 1/4! )
= ⁷C₃* ( 4!/2! - 4/3! + 4!/4! )
=⁷C₃* ( 12 - 4 + 1)
= ⁷C₃* 9
⁷C₃* 9 + ⁷C₅* 44 + ⁷C₁* 265
⁷C₃* 9 = 35 * 9 = 315
⁶C₂ * 21 = 15*21 = 315
⁷C₁ = ⁷C₆
⁷C₃* 9 + ⁷C₅* 44 + ⁷C₁* 265 = ⁶C₂ * 21 + ⁷C₅* 44 + ⁷C₆ * 265
option A & D are correct
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