Practice – Permutations & Combinations

1.  $3r+r+3=19\Rightarrow r=4$

2.  $n=6+4=10. \left(a\right)120 \left(b\right)66$

3.  $k^2–k=2+4\Rightarrow k=3$

4.  $C_9^{16}–C_7^{16}=0$

5. Total-(no. beginning with1)-(beginning with21)

$5!–4!–3!=90$

6. form numbers greater than 1000

$4!–3!$ (3! , no. beginnig with 0)=18.

7.  ${}^{6}P_3–{}^{5}P_2$ (no. beginning with 0)=100

8. 4 odd places and 4 odd nos.  $\times$ 3 even places (2nd, 4th, 6th)

= $\frac{4!}{2!2!}\times \frac{3!}{2!}=18$

9. there is only 1 way in which all lamps will be switched off

10. Two husbands can be selected out of 6 in  ${}^{6}C_2=15$ ways

excluding the wives of above 2 husbands, 2 ladies can be selected in = ${}^{4}C_2=6$ ways

so, number of ways = $15\times 6=90$

The above males and female can change/choose their partner in 2 ways, so total no. of ways $=90\times 2=180$