Name: Number of Onto Functions
Uploaded: 2019-07-19
Duration: 9 min 3 s
Description: In this video, I'll show you how to prove the number of onto functions formula. This will prove handy for a better understanding of this concept.

Question

How many $6$ digits odd numbers greater than $600000$ can be formed from the digits $5, 6, 7, 8, 9, 0$ if

(a) repetitions are not allowed?

(b) repetitions are allowed?

Accepted Answer

Since we only need odd numbers, then unit's digit can be either of $5, 7, 9$ only and for such numbers greater than $600000$ , first digit must be either of $6, 7, 8, 9$ only.

(a)

Case-I: Numbers starting with $6$ : The unit's digit can be filled in $3$ ways ( $5, 7$ or $9$ ) and remaining $4$ places can be filled by remaining $4$ digits in ${}^{4}P_{4} = 4! = 24$ ways. So, total numbers in this case $= 24 \times 3 = 72$

Case-II: Numbers starting with $7$ : The unit's digit can be filled in $2$ ways ( $5$ or $9$ ) and remaining $4$ places can be filled by remaining $4$ digits in ${}^{4}P_{4} = 4! = 24$ ways. So, total numbers in this case $= 24 \times 2 = 48$

Case-III: Numbers starting with $8$ : The unit's digit can be filled in $3$ ways ( $5, 7$ or $9$ ) and remaining $4$ places can be filled by remaining $4$ digits in ${}^{4}P_{4} = 4! = 24$ ways. So, total numbers in this case $= 24 \times 3 = 72$

Case-IV: Numbers starting with $9$ : The unit's digit can be filled in $2$ ways ( $5$ or $7$ ) and remaining $4$ places can be filled by remaining $4$ digits in ${}^{4}P_{4} = 4! = 24$ ways. So, total numbers in this case $= 24 \times 2 = 48$

So, total required number of such numbers $= 72 + 48 + 72 + 48 = 240$

(b)

Since unit's digit can be either of $5, 7, 9$ only, so unit's digit can be filled in ${}^{3}C_{1} = 3$ ways and for such numbers greater than $600000$ , first digit must be either of $6, 7, 8, 9$ only i.e. first digit can be filled in ${}^{4}C_{1} = 4$ ways. Each of the remaining digits can be filled in ${}^{6}C_{1} = 6$ ways. So, by principle of multiplication, total such numbers when repetitions are allowed $= 4 \times 6 \times 6 \times 6 \times 6 \times 3 = 15552$

How many $6$ digits odd numbers greater than $600000$ can be formed from the digits $5, 6, 7, 8, 9, 0$ if

(a) repetitions are not allowed?

(b) repetitions are allowed?

Important Questions on Permutation and Combination

Learn and Prepare for any exam you want !

How many 6 digits odd numbers greater than 600000 can be formed from the digits 5,6,7,8,9,0 if (a) repetitions are not allowed? (b) repetitions are allowed?

Important Questions on Permutation and Combination

Learn and Prepare for any exam you want !

How many $6$ digits odd numbers greater than $600000$ can be formed from the digits $5, 6, 7, 8, 9, 0$ if

(a) repetitions are not allowed?

(b) repetitions are allowed?