In how many ways 6 letters can be put in 6 addressed envelopes so that only 3 of them will go in the correct envelopes?

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

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

Open in App

Solution

The correct option is A
40

Here, we have to select 3 letters out of 6
Number of ways is $^{6} C_{3}$
We should de arrange the 3 letters.
Required number of ways = $^{6} C_{3} \times 3! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!})$
=40

Suggest Corrections

Similar questions

Q. In how many ways 6 letters can be put in 6 addressed envelopes so that only 5 of them will go in the correct envelopes.

6 letters are to be placed in 6 addressed envelopes. Each letter is assigned for an envelope. In how many ways these letters can be put in these envelopes such that exactly one letter goes in the correct envelope.

Q. There are 4 different letters and 4 addressed envelopes. In how many ways can the letters be put in the envelopes so that at least one letter goes to the correct address?

Q. There are four different letters and four addressed envelopes, in how many ways can the letter be put in the envelopes so atleast one letter goes into the correct envelope

The number of ways to put 6 letters in 6 addressed envelopes so that all are in wrong envelopes?

Join BYJU'S Learning Program

In how many ways 6 letters can be put in 6 addressed envelopes so that only 3 of them will go in the correct envelopes?

The correct option is A 40 Here, we have to select 3 letters out of 6 Number of ways is 6C3 We should de arrange the 3 letters. Required number of ways = 6C3×3!(1−11!+12!−13!) =40

The correct option is A
40

Here, we have to select 3 letters out of 6
Number of ways is $^{6} C_{3}$
We should de arrange the 3 letters.
Required number of ways = $^{6} C_{3} \times 3! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!})$
=40