In how many ways can a person go from point A to point B if he can travel only to the right or upward along the lines (Grid Problem)?

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $20$
To reach B from A he has to take three right steps and three upward steps no matter what he takes first . So we consider this problem as arrangement of 2 distinct objects(i.e right step and upward step ) 3 each .

So answer is $\frac{6!}{3! \times 3!} = 20$

Suggest Corrections

Similar questions

A person can only move upwards or towards right. In how many ways can he travel from A to C via B

Q. A person can only move upwards or towards right. In how many ways can he travel from A to B via C (refer diagram)?

Q. How many possible routes are present from points A to B if a person can only move to the right or downwards in straight lines?

In the adjoining figure, the lines represent on-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

Join BYJU'S Learning Program

In how many ways can a person go from point A to point B if he can travel only to the right or upward along the lines (Grid Problem)?

The correct option is B 20To reach B from A he has to take three right steps and three upward steps no matter what he takes first . So we consider this problem as arrangement of 2 distinct objects(i.e right step and upward step ) 3 each .So answer is 6!3!×3!=20

The correct option is B $20$
To reach B from A he has to take three right steps and three upward steps no matter what he takes first . So we consider this problem as arrangement of 2 distinct objects(i.e right step and upward step ) 3 each .

So answer is $\frac{6!}{3! \times 3!} = 20$